# Are laws of nature really the same in all reference frames?

1. Sep 24, 2011

### Bjarne

Let’ say; “A” can see and measure a stone falls to the earth let’s say 10 meter per 1 Earth-second.
“B” lives at Mercury and can see the same thing.
But “B” would do not see the exactly the same, because seen from “B’s” viewpoint time / distance is not the same as for “A”.
Let us say time at Mercury would tick half so fast compared to a clock at the Earth.

B would not agree it took the stone 1 second to move 10 meter – but have seen that the stone only was moving ½ -mercury second.
B will therefore also not see the stone falling 10 meter (as A saw it was falling in one Earth-second), but only that the stones was falling 5 Mercury-meter.

It must matter whether the stone was falling 10 meter (from A’s viewpoint) in a certain period, - or only 5 meter (B’s viewpoint), - So the problem is now, how can all laws of nature be the same for all observers.

If distances not are changing proportional the same rate as time, - A and B would not agree of the speed of light. Hence distance always must change proportional with time, right?.

Both A and B would therefore observe the “same” speed of the stone, - even though a process on Mercury would take relative double so long time measured with a Earth-clock.

It is simple math to understand that the speed of the stones anyway “seems” to be the “same” for both of A+B, - but in fact it is not, simply because time is different, and distance too.

For example if we on Earth (A) see a photon traveling from the Moon to the Earth, and it take 1 second, - the same event (according to the example) would seen from Mercury only take ½ second.
But because distance seen from the Mercury viewpoint (between the Moon and the Earth) is only the half compared to the earth viewpoint, - a photon would hence after one Mercury-second have travelled the double distance meassured 1 second, - with a Earth-clock.
Which mean that after 1 Mercury second the photon must have traveled 600,000 Earth-km measured in 2 second with a Earth clock. (Since 2 Mercury-second = 1 Earth-seconds)

After 1 orbit of the Milkyway, a clock at Mercury (B) would REALLY have “lost” 6 years compared to a clock at the Earth (A).

The point is that when time/distance not is the same for A and B, how can the laws expressed by Newtonian and Keplerian equations be the same everywhere.

At least the gravity constant “G” seems to must be adjusted all the time, since distance is changing all the time.
Otherwise the result of gravity will not be right by our feeds compared to ours noses.
How can a person that not share ours time-distances share (our) gravity constant (G) ?

For example;
A person living at mercury and another at the Earth could never agree about the distance - our Sun - travels the MilkyWay, - simple because time is not the same these two places.
Evidence is atomic clock wouldn’t lie on these two planets.

When 2 such observers cannot agree about distances /radius/ diameter of the Milkyway, - how is it possible for both to use the excact same gravity equations ?

If we exaggerate and say that a clock on Mercury ticks half so fast as on Earth, - this would mean that after 1 orbit of the Milkyway we on earth have travels 377,000 Light years, but a person living at Mercury would say the orbit only is the half.

Therefore 2 such observers must also get two different result of how strong gravity of the Milky way really is ?

How can we then say that the laws of Newtonian/Keperian gravity are the same for both observers?

Last edited: Sep 24, 2011
2. Sep 24, 2011

### ghwellsjr

When we say the same laws apply, we're talking about formulas but the numbers we plug into those formulas and the results can be
different on each planet, except everone will measure the same value for the speed of light.

3. Sep 25, 2011

### Bjarne

The Point is that the gravity constant "G" cannot be the same, at different planets (etc), even not the same by your feed as by your nose, since time and distance not are the same.

I you would live on Mercury you would also measure the "same" speed of every motion included light, but when you would compared such speeds, with how an observer on earth would see the same event these are not the same speed because distances of Mercury are shorter.

For example according to the example mentioned above, - after 1 Mercury-second the photon must have traveled 300,000 Mercury-km - but at the same time (period) it have travel 600,000 Earth-km
This is because 2 second measurement with a Earth clock is 1 Mercury-second measurement with a Mercury clock .

The observer on Mercury would hence say 600,000 Earth km is only 300,000 Mercury-km, and therefore in fact the photon must REALLY be moving double so far measured with a Earth-Meter stick compared to a Mercury meter-stick. ( but still the "same" speed/ distance seen from the perspective of both observers (without comparing).

This must mean that the observer plays the "primary role", - the Universe plays a secondary role.
Or that; - the Universe is like the eyes (clocks) see it, and not opposite.

Last edited: Sep 25, 2011
4. Sep 25, 2011

### ghwellsjr

You are right, the gravity constant, "g", is not the same on different planets, it's not even a constant on earth but changes slightly from location to location due to differences in the mass density of the earth at different locations and due to changes in elevation, but the gravitational constant, "G", does not change due to any local considerations. I would suggest you look up the articles in wikipedia on these two "constants" if you want to learn about them. But these issues have nothing to do with different laws at these different locations.

You have expressed concerns about the effects of gravity on measurements which is a very complex subject and requires an understanding of General Relativity. I would suggest that you focus first on understanding Special Relativity because that is so much easier and I think that once you can see how different observers traveling with respect to each other (no gravity involved) can both measure the speed of light to be the same, even though they each measure the other one's clocks and rulers to be different than their own. That, after all, is what this forum is for, understanding relativity. Are you interested in learning and understanding Special Relativity?

In the meantime, I'd like you to think about your comment regarding seeing a photon traveling from the moon to the earth and I would like to ask you, how can anyone see a photon? This has bearing on your last comment:

This must mean that the observer plays the "primary role", - the Universe plays a secondary role.
Or that; - the Universe is like the eyes (clocks) see it, and not opposite.​

5. Sep 25, 2011

### Bjarne

"g" is not part of the question.

I geuss you would write that.

The laws are the same, but but G can’t be, - it seems to mathematical impossible.

I can't see why it should be "so complex"
The essence is 2 clocks are ticking differently, -; this is all we so fare need to know.
We have 2 clock and 2 different observers. One observer could be you, - the other a man and a clock on Mercury.
How long distance would the Sun travel seen from the perspective of these two observers (in 377,000 years / orbit the MW) - and wich distance is the “right distance” ?.

After 1 orbit of the Milkyway (MW) these 2 observers could impossible agree about what the radius / diameter circumference or the MW is.
If time is different and speed of light must be the "same” for all observers - left is only that distances not can be the same.
It seems to me to be a simple mathematic necessity.
So why make this simple event more difficult as is possible can be?

I believe I know a lot about it, - but still we only dealing with simple facts; - 2 observers/clocks ticking with different rate cannot agree how the distance is of the MW.

Doesn’t matter whether the reason to the different time rate is GR or SR, - distances cannot be the same, and hence the Keplerian and Newtonian laws of gravity , - yes are the same, - but G cannot be, - since distances is the main factor of gravity.

Don't take it literary..
The principle is what count.
When time not is the same distances can also not be, - and hence "G" can also not be.
On the one hand speed seems to be the same for both observer, - on the other hand, - so soon you compare how the distance difference (measured by 2 different relativistic observers) is , the observer with the slow clock must measure a shorter comparable distance, - in a certain period common for both.

So on the one hand, any observer will see (measure) light (a photon) travels 300,000 km/h, -but only because of distances not can be the same, - on the other hand that picture is wrong so soon you compare what have happen in a certain period.

I think we shall be carefully not to use relativity as a junkyard for things that not make sence, and allow export to such junkyard so soon something is “so complex” that probably nobody has understood it.

So fare I see this question, there MUST be a simple logical mathematical explanation, doesn’t, matter whether the clock ticks different due to gravity (GR) or fast motion (SR)..

“If you can't explain it simply, you don't understand it well enough”.
Albert Einstein

Last edited: Sep 25, 2011
6. Sep 25, 2011

### ghwellsjr

You have stated that you believe you know a lot about SR but yet you have said:
But while time dilates (gets longer), length contracts (gets shorter) so it's not proportional, it's an inverse relationship. Furthermore, it's only distance in certain directions that is contracted. How do you explain this?

Last edited: Sep 25, 2011
7. Sep 25, 2011

### bobc2

Bjarne,

Why not try approaching it with a simpler comparison. Have A and B initially together in the same inertial reference system (on earth, out in space away from everything..., you pick it), and each is furnished with a laboratory on separate identical space ships. They can perform any of the basic experiments historically used to describe our physical laws.

Now, they take off in opposite directions at relativistic speeds (it really doesn't matter whether they go at the same speeds relative to the original rest frame or not). Each one performs a number of different experiments and then return to their original rest frame to compare their results. Their results will be the same.

So, yes--the laws of physics are the same in all inertial reference frames.

Also, if you have them land on distant planets (ignoring the inhabitable environments) and perform experiments, they will come up with the same results such as Newton's law of gravitation; they will find that masses attract with a force inversely proportional to r^2 with the same Newton gravitation constant, G. They will both find F = ma. The ratio of e/m will be the same, etc.

They probably won't have any more luck with experiments aimed at unification of general relativity and quantum mechanics than anyone has had so far.

Last edited: Sep 25, 2011
8. Sep 26, 2011

### Bjarne

Once again, there are no reasons to limit this question to Special Relativity.
The fact is that a clock on board on Mercury really is ticking slower as one at Earth.
Hence an observer (a clock) on Mercury must have "lost" 6 years due to GENERAL relativity, (compared to a clock at the Earth) after one MW orbit of the solar system.

This give you 2 simple possibilities

1.)
An observer on Mercury must have seen the Sun (orbit the MW) faster as an observer at the Earth have observed the same event (faster as the 250 km/s - the speed we on Earth observe the Solar system (the Sun) is orbiting the MW) - I don't believe you can sell that to anyone because that would in the end violate that “c” always is the same.

2.)
That distance is not the same for both observers. - This is logic math since an observer on Mercury would see the Sun complete 1 orbit of the MW in less time (6 years less) as seen from the Earth.

Last edited: Sep 26, 2011
9. Sep 26, 2011

### ghwellsjr

The title of your thread is "Are laws of nature really the same in all reference frames?". You are referring to Einstein's first postulate for Special Relativity and it is true only for inertial reference frames. It is not true in accelerating frames which is what you have on the surface of planets and which is handled by General Relativity.

You have stated:
But it does matter because in SR, time dilation is reciprocal and in GR it is not. It's only the reference frames in SR where the laws of nature are the same. Under the influence of different gravity fields, the laws are different. To put it another way: a scientist inside a box in any inertial frame cannot determine which kind of an inertial frame he is in, they are all the same due to the reciprocal nature of time dilation (among other things). But if you put him in an accelerating box, or in a box on the surface of a planet, he will be able to tell the difference between the boxes because they can have different accelerations and the time dilation is not reciprocal between two boxes on different planets (or on the same planet at different elevations).

10. Sep 26, 2011

### A.T.

It's not that they are different, but rather that in GR inertial frames don't extend infinitely. So in GR the equivalence of inertial frames only applies to local experiments.

11. Sep 26, 2011

### zonde

Under relativity inertial frames are constructed so that physical laws stay the same.

Now you say that inertial frames are not extendable globally. To me this statement seems equivalent to the statement that physical laws are not the same globally.

12. Sep 26, 2011

### ghwellsjr

I thought I made it very clear in my post that I was addressing the non-inertial frames on the surfaces of planets which is what Bjarne is using in his argument against Einstein's first postulate of Special Relativity. How does your post help Bjarne?

13. Sep 26, 2011

### ghwellsjr

You've got it backwards: Under relativity, physical laws are constructed so that they stay the same in different inertial frames, if they are not already that way.

14. Sep 26, 2011

### Bjarne

Now you confuse me.
My concern is mainly if ALL laws of nature (equations) ALWAYS the same for all observer.
OR is (for example) gravity - and here I mean G (the gravity constant) an exception´.
Above you wrote; "G", does not change due to any local considerations"
I am now not sure what you really mean, - Is "G" ALWAY constant. - Yes or no ?

So is it true that G is not the same by your feed as by you nose ?
Yes of no ?
If the answer is no, then try to explain how 2 observers located in different space-time (caused by gravity) obviously not would be able to agree how long time one MW orbit take, - and therefore logical also not agree what the radius of the MW really is , - and therefore also not agree what G really is ? - What I mean is; - there are no common answers. G cannot be a constant - OR WHAT ?

I prefer first to finish the easy part (the GR part) before confusion the discussion with SR.
So fare according to this context I believe a clock (observer) on Mercury don't care why time ticks slower, - exactly as we don’t care why our clock is slower as a clock on Neptune, I mean how much time dilation is due to GR and SR - the point will still be is G ALWAYS constant ?

Yes or no.

Last edited: Sep 26, 2011
15. Sep 26, 2011

### A.T.

Yes, not in curved space time.
No, this not equivalent. You can go anywhere and perform a local experiment in an inertial frame there, and then you get the same results. That is what "physical laws are the same globally" means.

16. Sep 26, 2011

### JordanL

In other words, globally there is local consistency, at the cost of locally there being global consistency?

EDIT:

Actually, consistency is the wrong word. What I suppose I meant was that globally there is local continuity of physical laws, instead of locally there being global continuity of physical laws. That's, as far as I understand it, rather the entire point of SR.

17. Sep 26, 2011

### ghwellsjr

It's no wonder you are confused when you don't read my posts. Let me reiterate:
The effect of different gravity constants "g" on different planets will not cause scientists on those planets to arrive at different values of the gravitational constant "G". We have the same problem here on earth where g varies but you have to take that into consideration when calculating constants that are independent of your local situation. The value of "G" is not dependent on any particular local circumstance.
I have said G is constant, g is not constant.

Is the reason that you think GR is easier than SR because things like time dilation and length contraction are not reciprocal?

18. Sep 27, 2011

### Bjarne

We did not got to the point. I mean we have a missing link here.
Let us forget everything about SR and only consider the consequence of GR and hence time dilation caused by gravity for 2 observers orbiting the MilkyWay.

These 2 observers, one on the Earth and one on Mercury measure the orbit of the Sun round the Milkyway, by multiplying time and speed.

Now...
The observer on Earth will claim 1 orbit of the MilkyWay took exactly 1,19E16 Earth-second.
But the observer on Mercury would say that the orbit toke 194,000,000 s (6 earth-year) less according to his clock.

This time difference is only caused by GR...

[PLAIN]http://www.science27.com/forum/td2.jpg [Broken]
Time dilation on Mercury relative to the Earth = 0,000000016 s. ( due to GR)
Time per year = 60*60*24*365 = 31,153,000 s.
Orbit of the MW = 314,000 Light Years (at the periphery).
Earth-Time to orbit the MW (250 km/s) = 377,000,000 years. (at the periphery).
Total Earth-second to 1 MW orbit = 377,000,000 * 365*24*60*60 = 1,19E16 seconds.
Lost" of Mercury-time in years = 1,19E16 s. * 0,000000016 s. = 194,000,000 s.

So after 1 MikyWay orbit the clock on Mercury have "lost" 194,000,000 seconds ( = 6 earth-years) relative to a clock on the Earth.

Now here is the simple question.
Is the circumference of the Milkyway the same for these 2 observers, so long we only speak about the mentioned time dilation due to GR ?

If the answer is yes, - how is this possible when the rate of time not is the same ?

This only can mean a mathematical meltdown, right ?

I mean since speed multiplied with the time one orbit takes must result to = Distances.

The only possible outcome I can see is that distance not can be the same, - so how can the law of gravity / the equations (and G) be the same?

I don’t understand why it is necessary to complicate that simple question - more as necessary, - .

Distances cannot be the same so far I can understand this simple logic , - this is to me the only logical answer, - and this must mean G have a problem.

It is not enough to say this is wrong, I must know why it should be wrong.

• Is the time rate on Mercury, - the way a clock would count it, - only an illusion ?
• Is speed seen from a Mercury perspective different?
• Is Mass seen from a Mercury perspective not the same as on Earth
• Or WHAT , - if not distance ?

Last edited by a moderator: May 5, 2017
19. Sep 27, 2011

### JordanL

All this example shows is that the same measurement if we attempt it from different reference frames will yield different results, which is rather the point of them, isn't it? Distances, speeds and times are not absolute, they are relative... which is why it is called relativity. You cannot say the distance is shorter, because that has no meaning. It only appears so from particular reference frames.

20. Sep 27, 2011

### Bjarne

You can say that distance (for exsample near the Sun) mathematical and logical must be relative shorter, than for example here.

21. Sep 27, 2011

### zonde

It should be that G is numerically different.
G is global constant. And if it is global constant then locally it should be different at different gravitational potentials due to time dilation.
Or maybe it is more reasonable to find out if it's GM product that's different or not and then talk about G and M separately.

Gravitational time dilation should be real effect as you can observe it in static setup.
I suppose that observations are consistent with distances being the same.

Anyways question seems quite interesting.

EDIT: Ah, but certainly speed is different from Mercury perspective due to time dilation. So the question is if it is enough to make picture consistent.

Last edited: Sep 27, 2011
22. Sep 27, 2011

### zonde

If you perform Cavendish experiment at different gravitational potentials will it give the same results?
Hmm, only it is performed in non-inertial frame as almost all of the physics experiments. Then maybe discussion about inertial reference frames is not very useful?

23. Sep 27, 2011

### ghwellsjr

OK, will do.
No.
They're not.

• No.
Yes.
It's not the same.
Everything is different.

Last edited by a moderator: May 5, 2017
24. Sep 28, 2011

### Bjarne

Bjarne
So when we measure the speed of the Sun to travel 250 km/s, - an observer on Mercury (or on the surface of the Sun) would measure it to be a little more than this? - is this what you saying ? If so Why ?

Now imaging the same 2 observers measure the speed of a photon, we on Earth would measure it to be; “c” - would an observer on Mercury also measure that to be faster?

Can the speed difference, - significant or almost completely solve the "mathematical meltdown" described above, - or is this value almost irrelevant in this context?

Does the speed difference, - have anything to do with the cause of the the "mathematical meltdown" described above ?

Bjarne
I mean is let’s say the mass of the Sun is exactly 2E30 Kg.
Are you saying an observer on Mercury not would agree ?
If so, - why?
What about the mass of the Milkyway? - would the 2 observers also disagree ?

Can the mass difference significant contribute to solve the "mathematical meltdown" described above, - or is it almost irrelevant in this context?

Does the mass difference, - have anything to do with the cause of the the "mathematical meltdown" described above ?

What more as (time) speed and mass is different ?

Last edited: Sep 28, 2011
25. Sep 28, 2011

### zonde

Let's try to put it down in less chaotic way.

As two observers observe the same sequence of physical events the only thing they can change is their representation of this sequence i.e. coordinate system.
Relative time dilation between two observers is real as we can establish delay in sequence of signals with static distance. So first of all we have different scale for time dimension for two observers.
Distances should be the same as speed of light locally is changing in a way that is consistent with unchanging distances.
Result of this is that orbital speed around MW for observer on Mercury is faster by the same factor as time is delayed.

Now we calculate GM-product μ using the same formula for both observers.
$$v_m^2=\frac{\mu_m}{r}$$
$$v_e^2=\frac{\mu_e}{r}$$
As speed is faster for Mercury observer but distances are the same for both observers we have that GM-product is bigger by that speed scaling factor squared. As GM-product have dimensions of time squared in denominator it seems that we have consistent picture so far.

Now if we assume that G is the same for both observers then mass unit for Mercury observer should be smaller by speed scaling factor squared.
This seems plausible as lowering mass in gravitational potential should convert part of the rest mass into kinetic energy.

Does this reasoning seems fine?