Proving the Universality of Physical Laws: A Challenge for Kev

[Myslius]

How do you prove that the laws of physics are the same in all inertial frames of reference?

 

[JustinLevy]

How do you prove that the laws of physics are the same in all inertial frames of reference?

A quick comment first: the laws of physics are the same in all inertial frames of reference of the same handedness.
http://en.wikipedia.org/wiki/Inertial_frame_of_reference

Now to answer your question. The rough approach is:
-- Define an inertial coordinate system. Deduce the most general coordinate transformations between such frames allowed by the definition. Now choose an inertial coordinate system to map out your surroundings and experimentally determine the laws of physics in that frame. Now mathematically check if applying the general inertial coordinate transformation to these laws of physics leaves the laws the same.


Historically, this actually happenned the other way around. We had a rough idea of an inertial frame, but found the physics had a different symmetry. This caused a lot of confusion at first, but eventually realized our approach to inertial coordinate systems was incorrect (the Newtonian concept of absolute time). Now that we have the concept of a spacetime, most definitions of an inertial frame essentialy are just defining what the metric for flat spacetime should be in such a coordinate system. We can then deduce the most general coordinate transformations preserving those metric components (lorentz boosts, rotations, translations, etc.) and check if the physics we measure is invariant under these coordinant transformations.

So when you here about an experimentalist seaching for "lorentz violating dynamics/processes/etc", they are testing the principle you asked about.

 

[nismaratwork]

Length contraction, Time dilation, and other Relativistic effects go a long way towards to showing the invariance of the laws of physics across IRFs. Essentially, every proof for SR/GR is a proof of that postulate.

 

[jcsd]

You don't is the answer, it's a postulate. Special relativity is designed to preserve the laws of physics in different inertial frames, so any argument that it proves this statement is circular. If you like an inertial frame by definition is a frame belonging to a certain class of frames where the laws of physics and the speed of light are invariant (and constant) under transfomations between frames belonging to this class.

Postulates are generally speaking inferred emprically, i.e. from experiment. The success of special relativity for describing certain physical phenomena can be taken as 'proof' of it's postulates.

You could use a different set of postulates which allow you to derive the postulates of special relativity. I still think that's a bit circular as essentiallly you're still setting out to create a theory which includes the postulates of SR.

 

[JesseM]

You don't is the answer, it's a postulate.

But it would certainly be possible to falsify the postulate experimentally, so it is testable in that sense. Basically I think the idea is that you can determine the correct equations that govern the dynamical behavior of particles/fields in the lab frame (like finding that charged particles obey Maxwell's laws in the lab-frame), and then if the equations are written in terms of x,y,z,t, you can see what happens when you use the Lorentz transformation to perform a substitution in your equations:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2)
y' = y
z' = z
(with gamma = 1/sqrt(1 - v^2/c^2))

If the equations after the substitution can be reduced to a form that looks identical to the original equations before the substitution (but with x' in place of x, t' in place of t, etc.), then the equations are Lorentz-invariant. It's logically possible that we might instead discover that the equations that give correct predictions in the lab frame do not have this property, which would falsify relativity unless we could show that the equations could be viewed as approximations to some other Lorentz-invariant equations which were consistent with all our experimental results.

 

[nismaratwork]

As I said, support for Relativity = support for its postulates, but theories aren't "proven" in physics; they're used, improved, and eventually discarded. I haven't seen any wrong answers to the OP's question in this thread, just differing approaches.

 

[bcrowell]

How do you prove that the laws of physics are the same in all inertial frames of reference?

To prove that it's true, you would need to start with a set of assumptions and then carry out some logical reasoning. What set of assumptions do you have in mind?

To prove that it's not true, you would need to find an experiment that comes out different when you do it in different inertial frames of reference. For example, if the Michelson-Morley experiment had come out with a positive result instead of a negative one, it would have disproved your statement.

The two possibilities --- proof and disproof --- are totally asymmetrical. One experiment can disprove a theory. No number of experiments can prove a theory.

 

[Myslius]

Can I choose FOR freely?
An experiment, let's take a satellite orbiting around the Earth as FOR. The Earth moves so time on Earth should go slower (SR, time dilation).
In reality time goes slower for satellites.

 

[starthaus]

Can I choose FOR freely?
An experiment, let's take a satellite orbiting around the Earth as FOR. The Earth moves so time on Earth should go slower (SR, time dilation).
In reality time goes slower for satellites.

This is due to the difference in gravitational potential. You need to take it in consideration in your calculations. See http://relativity.livingreviews.org/Articles/lrr-2003-1/ .

 

[Ich]

Myslius, please explain what you think "inertial" means in a SR context.

 

[atyy]

A quick comment first: the laws of physics are the same in all inertial frames of reference of the same handedness.
http://en.wikipedia.org/wiki/Inertial_frame_of_reference

I guess this refers to parity violation?

Classically it makes no difference, does it?

 

[Myslius]

Myslius, please explain what you think "inertial" means in a SR context.

My understanding about "inertial" was wrong. Inertial in SR context means that is it not affected by any force, and goes at a constant speed in the straight line.
Satellite is affected by Earth's gravity. So it can't be FOR. Right?

Actually, none object with mass can be inertial FOR.

 

[Myslius]

What observations conflict with relativity? As far as i know, relativity does not work at extremums: black holes, beggining of the big bang etc.

 

[Ich]

Satellite is affected by Earth's gravity. So it can't be FOR. Right?

In the context of SR, right.

Actually, none object with mass can be inertial FOR.

In this context, you'd find not a single exact inertial frame. But there are many situations where you can simply neglect the deviations, and where SR calculations are sufficient.
For example, the lab frame in a circular particle accelerator is inertial enough for all practical purposes.
The frame of a particle there can be approximated by an inertial frame, but only for a few meters - as long as their path looks almost straight.
Things that go in circles definitely do not qualify as inertial in SR.

 

[Passionflower]

In reality time goes slower for satellites.

Satellite clocks go faster not slower with respect to a clock on earth.

 

[JesseM]

Satellite clocks go faster not slower with respect to a clock on earth.

If you're considering only gravitational time dilation that'd be true, but for satellites in orbit around Earth I think velocity-based time dilation would have a larger effect, causing them to have elapsed less time on each successive orbit when they pass near an Earth-based clock (and if you're imagining a purely SR analysis of satellites where they are traveling in a circle in flat spacetime, which is what Ich was talking about with the comment 'In the context of SR', there would be no gravitational time dilation)

 

[Passionflower]

If you're considering only gravitational time dilation that'd be true, but for satellites in orbit around Earth I think velocity-based time dilation would have a larger effect, causing them to have elapsed less time on each successive orbit when they pass near an Earth-based clock (and if you're imagining a purely SR analysis of satellites where they are traveling in a circle in flat spacetime, which is what Ich was talking about with the comment 'In the context of SR', there would be no gravitational time dilation)

That is not my impression, I thought that the gravitational "part" was stronger than the SR "part".

But we should consult the literature. I have a paper by Richard Shiffman with the exact calculations that seem to agree with me. But I am not sure if this paper is published in a serious magazine and peer reviewed. Wikipedia also seems to agree with me but Wikipedia cannot always be relied on. There is Neil Ashby's document in Living Reviews but I browsed it and could not find any place where "the rubber meets the road" where it said unequivocally that one is slower or faster than the other one.

By the way, contexts or not, do you agree there is only one valid answer whether the clocks go faster of slower?

Edited: I checked the http://relativity.livingreviews.org/Articles/lrr-2003-1/" article again and I think you want to take a look at Eq. 35, the satellite clock goes faster.

 

[starthaus]

If you're considering only gravitational time dilation that'd be true, but for satellites in orbit around Earth I think velocity-based time dilation would have a larger effect,

No, the effect is about 6 times smaller.

causing them to have elapsed less time on each successive orbit when they pass near an Earth-based clock (and if you're imagining a purely SR analysis of satellites where they are traveling in a circle in flat spacetime, which is what Ich was talking about with the comment 'In the context of SR', there would be no gravitational time dilation)

The gravitational effects heavily dominate the effects due to relative motion by a factor of +46us/day vs -8us/day, so, the terrestrial clocks lag the satellite clocks by a net of +38us/day. See here..
In order to compensate for this effect, the frequency of the atomic clocks is adjusted down at launch, making it one of the most direct tests for relativistic time dilation.

 

[starthaus]

Edited: I checked the http://relativity.livingreviews.org/Articles/lrr-2003-1/" article again and I think you want to take a look at Eq. 35, the satellite clock goes faster.

Yes, you were correct all along, by 38us day. This is why the frequency is adjusted down at launch (to make it count the same amount of units of time as the terrestrial clocks) The gravitational effects dominate the speed effects by a factor of 6.

 

[Ich]

Yes, you were correct all along, by 38us day.

Neither JesseM nor Myslius were talking especially about GPS satellites. Satellite clocks "go slower" for low Earth orbits, like Space shuttles, ISS, and most satellites. Mostly communication satellites (incuding GPS) are in higher orbits, with "faster" clocks.

 

[JesseM]

That is not my impression, I thought that the gravitational "part" was stronger than the SR "part".

But we should consult the literature. I have a paper by Richard Shiffman with the exact calculations that seem to agree with me. But I am not sure if this paper is published in a serious magazine and peer reviewed. Wikipedia also seems to agree with me but Wikipedia cannot always be relied on. There is Neil Ashby's document in Living Reviews but I browsed it and could not find any place where "the rubber meets the road" where it said unequivocally that one is slower or faster than the other one.

By the way, contexts or not, do you agree there is only one valid answer whether the clocks go faster of slower?

Edited: I checked the http://relativity.livingreviews.org/Articles/lrr-2003-1/" article again and I think you want to take a look at Eq. 35, the satellite clock goes faster.

Yes, you're correct, I misremembered. In an earlier post I had used some equations posted by kev to figure out that for a circular orbit, velocity-based time dilation would only be larger than gravitational time dilation for an orbit less than double the Schwarzschild radius (assuming all the mass was concentrated at a radius smaller than this, which isn't true for the Earth).

Still, when discussing the subject with beginners it may be better to consider a simplified case where we treat an orbit as a circular path in flat spacetime, as Ich did. In that case there is only velocity-based time dilation, and a circular path is non-inertial so a clock on that path will elapse less time than a clock at rest relative to the center of the "orbit".

 

[starthaus]

Yes, you're correct, I misremembered. In an earlier post I had used some equations posted by kev to figure out that for a circular orbit, velocity-based time dilation would only be larger than gravitational time dilation for an orbit less than double the Schwarzschild radius (assuming all the mass was concentrated at a radius smaller than this, which isn't true for the Earth).

The above cannot possibly be right since it is falsified by experiment. In the Hafele-Keating experiment one of the atomic clocks lagged behind the ground clock while the other was ahead. So, something is wrong in kev's computations.

 

[JesseM]

The above cannot possibly be right since it is falsified by experiment. In the Hafele-Keating experiment one of the atomic clocks lagged behind the ground clock while the other was ahead. So, something is wrong in kev's computations.

kev's calculation was only for comparing proper time on a clock moving in a circular orbit with coordinate time in Schwarzschild coordinates (which matches up to proper time for a clock at infinity).

 

[starthaus]

kev's calculation was only for comparing proper time on a clock moving in a circular orbit with coordinate time in Schwarzschild coordinates (which matches up to proper time for a clock at infinity).

Even that seems wrong since $$\frac{d\tau}{dt}<1$$ for any value of the Schwarzschild coordinate $$r$$ for the case of objects in circular orbits around a massive object like the Earth.
If you point to the exact post, I will be able to point out the error in kev's calculations. Did you check kev's calculations for correctness?

 

[JesseM]

Even that seems wrong.

Why do you think it seems wrong? Do you understand that the clock on Earth is moving in Schwarzschild coordinates due to the Earth's rotation, and that the plane that flies in the direction opposite to the rotation of the Earth would have a smaller speed in Schwarzschild coordinates than the clock on the ground?

If you point to the exact post, I will be able to point out the error in kev's calculations.

See pervect's post here where he calculates time dilation as a function of coordinate velocity in Schwarzschild coordinates--kev's calculation is just a slight modification where he substitutes the local velocity (velocity in a locally inertial frame instantaneously at rest in Schwarzschild coordinates) for the Schwarzschild coordinate velocity, see this post or this one. I remember now that we already discussed pervect's calculations on this thread where you raised a lot of spurious objections and wouldn't give straight answers to the questions I and others asked you, let's not have a repeat of that please (I will only agree to discuss this again if you agree in advance to give definite answers to any questions asked of you).

 

[starthaus]

Why do you think it seems wrong? Do you understand that the clock on Earth is moving in Schwarzschild coordinates due to the Earth's rotation, and that the plane that flies in the direction opposite to the rotation of the Earth would have a smaller speed in Schwarzschild coordinates than the clock on the ground?

Look, I am not going to engage in this Q&A game with you again. I know GR quite well and it takes one line of calculations to show that $$\frac{d\tau}{dt}<1$$ for any r.

See pervect's post here where he calculates time dilation as a function of coordinate velocity in Schwarzschild coordinates

I looked, pervect is using an incorrect metric, his formula for $$g_{tt}$$ is incorrect.
If you want, I can do the correct one-line calculation.

--kev's calculation is just a slight modification where he substitutes the local velocity (velocity in a locally inertial frame instantaneously at rest in Schwarzschild coordinates) for the Schwarzschild coordinate velocity, see this post or this one

I looked. kev's calculation has two mistakes:

1. He uses the wrong metric , i.e. wrong $$g_{tt}$$

2. even worse, he does not calculate the time dilation for orbital motion

As I suspected, it is alll wrong.

I remember now that we already discussed pervect's calculations on this thread where you raised a lot of spurious objections and wouldn't give straight answers to the questions I and others asked you, let's not have a repeat of that please (I will only agree to discuss this again if you agree in advance to give definite answers to any questions asked of you).

That thread dealt with yet different errors in calculating time dilation for radial (not orbital) motion. No point in bringing into play, do you realize kev's errors in calculating time dilation for orbital motion?

 

[yuiop]

I looked. kev's calculation has two mistakes:

1. He uses the wrong metric , i.e. wrong $$g_{tt}$$

2. even worse, he does not calculate the time dilation for orbital motion

As I suspected, it is alll wrong.

1) I used the right metric and corrected the typo in the metric quoted by Pervect.
2) I do calculate the time dilation specifically for orbital motion in the thread quoted by Jesse here https://www.physicsforums.com/showthread.php?p=2690217#post2690217

The above cannot possibly be right since it is falsified by experiment. In the Hafele-Keating experiment one of the atomic clocks lagged behind the ground clock while the other was ahead. So, something is wrong in kev's computations.

You do realize that the Hafele-Keating experiment used aircraft? You do realize that while aircraft can fly at any speed they like, that satellites are constrained to orbit at one speed for a circular orbit of given radius?

The equation for the time dilation of a satellite with natural motion was given by me in the other thread as:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{GM}{(Rc^2-2GM)}\right)\left(1-\frac{2GM}{Rc^2}\right)}$$

where T' is the proper time of the satellite clock and T is the time recoded by the Schwarzschild observer at infinity. The only variables are the mass of the massive gravitational body that is being orbited and the orbital radius because those factors determine the orbital velocity. As Jesse quite correctly observed, the contribution of the altitude related time dilation is greater than the contribution of the velocity related time dilation factor for any orbit that is greater than R=4Gm/c^2 which applies to any circular Earth orbit. It can also be noted that increased orbital radius causes reduced tangential velocity of the orbiting satellite and reduced gravitational time dilation and both effects speed up the satellite clock with increased altitude. The upshot is that the higher the satellite orbit, the faster the clock runs, relative to clocks in lower orbit, as Ich said.

If you need to calculate the time dilation of a clock with arbitrary motion, such as clocks in aircraft, helicopters or towers, then you need to use the more general equation that I gave in the same post which is:

$$T '/T = \sqrt{1-v^2/c^2}\sqrt{1-2GM/(Rc^2)}$$

If you wish to compare the clock rate T1 of a clock at R1 with velocity V1 with the clock rate T2 of a clock at R2 with velocity V2, then you can use:

$$\frac{T_1}{T_2} = \sqrt{\left(\frac{c^2-v_1^2}{c^2-v_2^2}\right) \, \left(\frac{1-\frac{2GM}{r_1c^2}}{1-\frac{2GM}{r_2c^2}\right) }$$

All you have to do is select the appropriate equation for the situation at hand.

For an oblate spinning Earth, the bulge causes a clock at sea level at the equator to be further away from the centre of the Earth than clocks at sea level at the poles, but the increased clock rate due to increased radius is exactly canceled out by the reduced clock rate due to increased rotational tangential velocity at the equator.

 

[starthaus]

1) I used the right metric and corrected the typo in the metric quoted by Pervect.
2) I do calculate the time dilation specifically for orbital motion in the thread quoted by Jesse here https://www.physicsforums.com/showthread.php?p=2690217#post2690217
You do realize that the Hafele-Keating experiment used aircraft? You do realize that while aircraft can fly at any speed they like, that satellites are constrained to orbit at one speed for a circular orbit of given radius?

And this is relevant why? Does the underlying physics change if you switch from satellites to planes? (Hint no, the results of the GPS are in perfect agreement with the results of the Haefele-Keating experiment)

The equation for the time dilation of a satellite with natural motion was given by me in the other thread as:

$$\frac{T '}{T} = \sqrt{\left(1-\frac{GM}{(Rc^2-2GM)}\right)\left(1-\frac{2GM}{Rc^2}\right)}$$

where T' is the proper time of the satellite clock and T is the time recoded by the Schwarzschild observer at infinity.

Unfortunately the above equation is wrong.

 

[nismaratwork]

Starthaus, Kev... why do you two always end in an argument about the minutiae of Kev's equations? This is waaaaaay past anything the OP was asking.

 

[Passionflower]

Starthaus, Kev... why do you two always end in an argument about the minutiae of Kev's equations? This is waaaaaay past anything the OP was asking.

A scientific formula is right or wrong there is no middle way. I certainly would appreciate it if my formulas or calculations were found wrong, that's the way to learn, you apply, make mistakes, and hopefully someone else takes the trouble of telling you you are right or wrong.

To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth. The time ratio is:

$$\frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}$$

In geometric units it is even simpler:

$$\frac{\tau_S}{\tau_E} = 1+\frac{M}{R} - \frac{3M}{2r}$$

G = 6.6726E-11
M = 5.9742E+24 kg (0.004435407 m)
R = 6378000.1 m

So the clock in the satellite is faster if $r >\frac{3}{2}R > 3189 km$ otherwise it is slower.

 

[starthaus]

A scientific formula is right or wrong there is no middle way. I certainly would appreciate it if my formulas or calculations were found wrong, that's the way to learn, you apply, make mistakes, and hopefully someone else takes the trouble of telling you you are right or wrong.

To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth. The time ratio is:

$$\frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}$$

In geometric units it is even simpler:

$$\frac{\tau_S}{\tau_E} = 1+\frac{M}{R} - \frac{3M}{2r}$$

G = 6.6726E-11
M = 5.9742E+24 kg (0.004435407 m)
R = 6378000.1 m

So the clock in the satellite is faster if $r >\frac{3}{2}R > 3189 km$ otherwise it is slower.

Yes, this is correct. Here is a more precise (but with same results) calculation:

Start with the Schwarzschild metric for $$dr=d \theta =0$$

$$(cd\tau)^2=\alpha (cdt)^2-(r d\phi)^2$$

valid for your satellite

and:

the Schwarzschild metric for $$dr=d \theta =d \phi=0$$

$$(cd\tau)^2=\alpha (cdt)^2$$

valid at the pole

where $$\alpha=1-2m/r$$

and $$m=\frac{GM}{c^2}$$

$$\frac{d \tau}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}<1$$ for any r
The above gives:

$$\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}$$

So:

$$\frac{d \tau_s}{d \tau_p}>1$$ if

$$2m/r+(r \omega/c)^2<2m/R$$

But, from the Kepler's law, it can be shown that :

$$(r \omega/c)^2=m/r$$

so, the above becomes:

$$3m/r<2m/R$$ i.e., your condition $$r>\frac{3R}{2}$$

The only problem I see with your proof is that : $$\frac{3R}{2}=3*6400/2=9600km$$ :-)

Now, GPS satellites are moving at about 20,000km above the Earth, so , if it weren't for the frequency precompensation at launch, their clocks would be faster than the ones left on Earth.

 

[Passionflower]

The only problem I see with your proof is that : $$\frac{3R}{2}=3*6400/2=9600km$$ :-)

Oops, you are right:

$$r-R > 3189 km$$

 

[starthaus]

Oops, you are right:

$$r-R > 3189 km$$

Out of curiosity, how did you arrive to your starting formula? As you can see, I had to go through a modest derivation.

 

[Passionflower]

Out of curiosity, how did you arrive to your starting formula? As you can see, I had to go through a modest derivation.

"General Relativity An Introduction For Physicists"
Hobson, Efstathious, Lasenby (Cambridge 2006)

Exercise 7.7

 

[starthaus]

"General Relativity An Introduction For Physicists"
Hobson, Efstathious, Lasenby (Cambridge 2006)

Exercise 7.7

OK, I don't have the book, I derived mine from scratch.