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Posted (edited)

The gravitational term is GM/rc2 and the kinetic term is v2/2c2 (using and expansion ignoring higher orders of c)

The first term is correct, the second term isn't quite correct. It is missing a [math]\frac{1}{1-r_s/r}[/math] (and so does the wiki page) . You are missing the third term, the one that quantifies the rotational motion, [math]\frac{r^2\omega^2}{c^2}[/math]. Technically, there can be even a fourth term, if the motion is not planar.

 

The correct response is that the equations are wrong, not that they can't be combined. And you link to the H-K experiment page, which shows/uses separate kinematic and gravitational terms, so while it's true these are all part of GR, it contradicts your claim.

My claim is that the solution doesn't "combine GR with SR", the whole solution is GR (as I have just shown). Please read carefully what I wrote.

Edited by xyzt
Posted

Multiplying the effects is a common mistake. It turns out that the effect is (heavily) non-linear , so you cannot resort to multiplication. The Schwarzschild solution provides you all the information necessary for performing the correct calculation. The Hafele-Keating experiment has been reprised with a much higher precision. Incidentally, the additive effect is also at the foundation of the GPS relativistic corrections.

The overall effect is derived from the Schwarzschild solution:

 

[math](c d \tau)^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2(d\theta)^2[/math]

 

giving:

 

[math]\tau=\int{\sqrt{1-r_s/r-(v/c)^2/(1-r_s/r)-(r/c)^2 \omega^2} dt}[/math]

 

For small values of [math]r_s, v/c, r \omega /c[/math] the above can be simplified to:

 

[math]\tau=\int{(1-0.5(r_s/r+(v/c)^2/(1-r_s/r)+(r/c)^2 \omega^2)) dt}[/math]

 

...recovering the explanation from the wiki article on Hafele-Keating.

 

 

 

Ok. But I must confess that I'm still confident that my simple reasoning is correct.

 

Let's try something even more simple:

 

First a clock is lovered into such a gravity well that it's frequency halves.

 

Then the clock is accelerated to coordinate velocity 0.5 c. This velocity happens to be the coordinate velocity at which a light beam propagates at that altitude.

 

So our clock is halted. Does GR say something different?

 

No, the fractional frequency terms add together.

 

 

 

Well, it sounds absurd that halving some rate in the bottom of a gravity well would be anything but halving said rate when observed from somewhere else.

 

A guy is driving a car in a gravity well. The rmp meter reads 1000 rpm. The fift gear is being used. Speed meter reads 100 mph.

 

Then said guy shifts to first gear. The rmp meter still reads 1000 rpm. Speed meter reads 10 mph.

 

Seen from far above the ratio of the two speeds must be 10. The number of cogs in the cogwheels dictates the speed ratio.

 

 

Now let's declare that the car is a clock. The shifting of gears caused a timedilation of the clock, by the factor of 0.1 , as observed from anywhere.

Posted

Using factors of 2 here is problematic, because the approximations in the formulae probably fail. The bottom line is that when using those equations, the kinematic and gravitational (and any other applicable) terms add to give the overall effect. A dilation value of 0.01 in each terms does not result in an overall dilation of 0.0001. It will be 0.02. You are using a value where the failure of your method is not obvious.


 

My claim is that the solution doesn't "combine GR with SR", the whole solution is GR (as I have just shown). Please read carefully what I wrote.

 

Which has nothing to do with my objection. Please read carefully what I wrote. There are two issues here — the source of the terms, and how they combine.

Posted (edited)

 

 

 

Ok. But I must confess that I'm still confident that my simple reasoning is correct.

 

Let's try something even more simple:

 

1. First a clock is lovered into such a gravity well that it's frequency halves.

 

2. Then the clock is accelerated to coordinate velocity 0.5 c. This velocity happens to be the coordinate velocity at which a light beam propagates at that altitude.

 

So our clock is halted. Does GR say something different?

 

1. The halving of the frequency happens when:

 

[math]\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}=\frac{1}{2}[/math]

 

i.e.

 

[math]\frac{4}{r_1}-\frac{1}{r_2}=\frac{3}{r_s}[/math]

 

2. In GR, the clock would "halt" if :

 

[math]1-\frac{r_s}{r}-\frac{(v/c)^2}{1-r_s/r}=0[/math]

 

This means :

 

[math]v=c(1-r_s/r)[/math] . This is way off from your [math]v=0.5c[/math]. Using intuition is a very bad way of tackling problems.

 

 

Which has nothing to do with my objection. Please read carefully what I wrote. There are two issues here — the source of the terms, and how they combine.

1. You got the "kinematic" term wrong, you are off by [math]\frac{1}{1-r_s/r}[/math]. In strong gravitational fields, the error is significant.

2. The terms do not really "combine" since the relationship is , in general, non-linear.

3. Only in very special cases (low speeds and weak gravitational fields) one can claim that the terms "combine".

4. Either way, this"combination" is not a combination of GR and SR, as claimed in the OP, the effect is purely GR since it is all derived from the GR metric.

Edited by xyzt
Posted

1. The halving of the frequency happens when:

 

[math]\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}=\frac{1}{2}[/math]

 

i.e.

 

[math]\frac{4}{r_1}-\frac{1}{r_2}=\frac{3}{r_s}[/math]

 

2. In GR, the clock would "halt" if :

 

[math]1-\frac{r_s}{r}-\frac{(v/c)^2}{1-r_s/r}=0[/math]

 

This means :

 

[math]v=c(1-r_s/r)[/math] . This is way off from your [math]v=0.5c[/math]. Using intuition is a very bad way of tackling problems.

1. You got the "kinematic" term wrong, you are off by [math]\frac{1}{1-r_s/r}[/math]. In strong gravitational fields, the error is significant.

2. The terms do not really "combine" since the relationship is , in general, non-linear.

3. Only in very special cases (low speeds and weak gravitational fields) one can claim that the terms "combine".

 

 

The example you gave was the Hafele-Keating experiment, for which none of these three objections apply. You even said the effects are additive in your post.

 

The fourth point has not been contended, and yet it continues to appear in the discussion. I think everyone gets that.

Posted

 

The example you gave was the Hafele-Keating experiment, for which none of these three objections apply. You even said the effects are additive in your post.

 

 

The effects are additive under certain conditions:

-low radial and tangential speeds

-low gravitational fields

You cannot claim that , in general, the effects are additive.

Posted (edited)

What about when you have a path that is not circular?

1. The halving of the frequency happens when:

[math]\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}=\frac{1}{2}[/math]

i.e.

[math]\frac{4}{r_1}-\frac{1}{r_2}=\frac{3}{r_s}[/math]

2. In GR, the clock would "halt" if :

[math]1-\frac{r_s}{r}-\frac{(v/c)^2}{1-r_s/r}=0[/math]

This means :

[math]v=c(1-r_s/r)[/math] . This is way off from your [math]v=0.5c[/math]. Using intuition is a very bad way of tackling problems.

1. You got the "kinematic" term wrong, you are off by [math]\frac{1}{1-r_s/r}[/math]. In strong gravitational fields, the error is significant.
2. The terms do not really "combine" since the relationship is , in general, non-linear.
3. Only in very special cases (low speeds and weak gravitational fields) one can claim that the terms "combine".
4. Either way, this"combination" is not a combination of GR and SR, as claimed in the OP, the effect is purely GR since it is all derived from the GR metric.

1. By combining them I meant combining the effects from gravity and velocity
2. The link you showed does not have the same equations as of what you posted
3. According to the expirment,

Total time dilation

\Tau = \Delta\tau_v + \Delta\tau_g + \Delta\tau_s

Velocity

\Delta\tau_v = - \frac{1}{2c^2} \sum_{i=1}^{k}v_i^2 \Delta t_i

Gravitation

\Delta\tau_g = \frac{g}{c^2} \sum_{i=1}^{k} (h_i - h_0) \Delta t_i

Sagnac effect

\Delta\tau_s = - \frac{\omega}{c^2} \sum_{i=1}^{k} R_i^2 \cos^2 \phi_i \Delta\lambda_i

Toffoto's approach seems logical, but according to the expirment, it is not true. But why do they add and not multiply?

Edited by Endercreeper01
Posted (edited)

 

2. The link you showed does not have the same equations as of what you posted

 

Actually, it does, since I wrote it as well. The only difference is that it shows how the effects are calculated on each segment of the trip in the experiment (the plane took off, reached altitude, slowed down, landed a few times, etc).

Edited by xyzt
Posted

Actually, it does, since I wrote it as well. The only difference is that it shows how the effects are calculated on each segment of the trip in the experiment (the plane took off, reached altitude, slowed down, landed a few times, etc).

What other arguments do you have?

Posted

What other arguments do you have?

The argument that the math is the same. The fact that you don't recognize this simple fact is irrelevant.

But why do they add and not multiply?

Because correct application of GR teaches you so.

Posted

The effects are additive under certain conditions:

-low radial and tangential speeds

-low gravitational fields

You cannot claim that , in general, the effects are additive.

 

A good thing you clarified this, because your earlier post just made a blanket statement.

 

GR makes a different prediction than the one you have, the changes are much more complicated than you think. A good synopsis can be found in the description of the Hafele-Keating experiment. The effects are additive, not multiplicative.

 

(emphasis added)

Posted (edited)

 

A good thing you clarified this, because your earlier post just made a blanket statement.

 

 

(emphasis added)

This is why I gave the derivation from first principles, to show the rigorous answer, rather than its components and to show the conditions under which the total effect is the sum of the components. These conditions are restricted to both weak gravitational fields and to low speeds wrt. light speed. I am glad that we are in clear now. Do you understand the other issues with your answer? You never acknowledged the issues, this is why I am asking.

Edited by xyzt

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