# Reference frame symmetry in Special Relativity

• B
• x-vision
In summary, the distance between the rocket and the star at t(0) is 10 light years, but the distance between the rocket and the star at t(0) when frozen would be the same in both frames.
x-vision
Hello, I have a couple of questions related to reference frames in Special Relativity.

Let's consider a rocket that is inertially moving towards a star with a relative velocity 0.9c.
I'd like to look at this example from both the rocket's and the star's perspectives.

In the reference frame of the rocket:
• The rocket is at rest and the star is moving towards the rocket.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the star will reach the rocket in 11.1 years.
• From the rocket's perspective, time is slowing down for the star, so only 4.8 years will have passed in the star's reference frame.
In the reference frame of the star:
• The star is at rest and the rocket is moving towards the star.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the rocket will reach the star in 11.1 years.
• From the star's perspective, time is slowing down for the rocket, so only 4.8 years will have passed in the rocket's reference frame.
I have calculated the 4.8 years interval using the time dilation formula:

• Is my math correct ;)
• Given that there is no acceleration involved in this example, can we safely assume that the two reference frames are fully symmetrical?
• When we switch the roles of "stationary" and "moving" between the star and the rocket, the proper distance between them doesn't change, right?
• The proper distance in this example is always in the reference frame of the stationary observer.
Thanks

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x-vision said:
At time t(0), the distance between the rocket and the star is 10 light years.

Only in one of the frames. You need to specify which one. It is not possible for the distance to be 10 light years on both frames, yet that's what your descriptions say; so your descriptions can't be correct as they stand.

x-vision said:
• When we switch the roles of "stationary" and "moving" between the star and the rocket, the proper distance between them doesn't change, right.
• The proper distance in this example is always in the reference frame of the stationary observer.

The term "proper distance" does not have a single unique meaning. (In that respect it is unlike the term "proper time", which does; that makes it unfortunate that the adjective "proper" is used in both cases, but that's the terminology we're stuck with in the SR literature.) You should ignore "proper distance" and focus on the distance in each frame without worrying about which one is the "proper" distance.

x-vision
x-vision said:
Let's consider a rocket that is inertially moving towards a star with a relative velocity 0.9c.
At time t(0), the distance between the rocket and the star is 10 light years.

This initial statement of the problem implies that the distance is 10 light years in the star's rest frame. I would recommend taking that as your starting problem specification. Then your calculations and statements about the star's reference frame are correct.

What you need to fix then is your calculations and statements about the rocket's reference frame. If we accept the specifications as given above, the first of your statements about the rocket's frame is correct, but the other two are wrong.

x-vision
PeterDonis said:
This initial statement of the problem implies that the distance is 10 light years in the star's rest frame. I would recommend taking that as your starting problem specification. Then your calculations and statements about the star's reference frame are correct.

What you need to fix then is your calculations and statements about the rocket's reference frame. If we accept the specifications as given above, the first of your statements about the rocket's frame is correct, but the other two are wrong.

Thanks. I have fixed my questions and specified that t(0) applies to both reference frames.
With this clarification, can we assume that the distance between the two reference frames stays the same when we switch the roles of "moving" and "stationary"?
If the clocks are synchronized at t(0), is this clarification still necessary?
Thanks again.

x-vision said:
have fixed my questions and specified that t(0) applies to both reference frames.

That doesn't fix anything.

x-vision said:
With this clarification, can we assume that the distance between the two reference frames stays the same when we switch the roles of "moving" and "stationary"?

No. Nothing you have done changes anything I posted previously.

x-vision said:
If the clocks are synchronized at t(0), is this clarification still necessary?

The clarification is wrong to begin with. See above.

What references have you used to learn SR?

PeterDonis said:
That doesn't fix anything.

I honestly don't understand your comment.

Let me explain where I'm coming from, so you can maybe point me to my blind spot.
Let's imagine that we can freeze the time on both systems at t(0).
I would think that at this time the distance between the rocket and the star would be the same, irrespective of the perspective: either from the the rocket, or from the star.
Is this assumption incorrect?

Thanks

x-vision said:
Is this assumption incorrect?

Yes. Time is relative; that means "freezing time" means different things in different frames. In particular, it means drawing different spacelike surfaces of constant time that are tilted with respect to each other.

Again, what references have you used to learn SR? Have you ever seen a spacetime diagram? Drawing a spacetime diagram should make it obvious why your assumption is incorrect.

PeterDonis said:
Again, what references have you used to learn SR? Have you ever seen a spacetime diagram? Drawing a spacetime diagram should make it obvious why your assumption is incorrect.
Thanks. Spacetime diagrams might be too advanced for me at this time.
I guess my question was more about the symmetry of reference frames.

In my mind, the distance between two bodies would be the same, irrespective of whether we look at it from the left-hand side or from the right-hand side, so to speak.
So, if we select either body as our stationary reference frame, we should see the same distance at time t(0)..
After that this distance will look different from the two reference frames, of course, but shouldn't it be the same at the starting point?

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x-vision said:
Spacetime diagrams might be too advanced for me at this time.

Have you seen the equations for a Lorentz transformation?

x-vision said:
I guess my question was more about the symmetry of reference frames.

The symmetry of reference frames does not mean what you appear to think it means.

x-vision said:
Or not?

Not. "The starting point" does not mean what you appear to think it means.

You appear to not understand the basics of SR. Once more: what references have you used to learn SR?

PeterDonis said:
Once more: what references have you used to learn SR?
Different articles on the Internet. Still finding my way.

The symmetry of reference frames does not mean what you appear to think it means.

Hmm. I thought that if the roles of "stationary" and "moving" are interchangeable, then the symmetry is implied. I guess not.

Thanks

x-vision said:
Different articles on the Internet. Still finding my way.
This is not a very good learning strategy. The internet is full of dubious information that may be misleading and/or wrong. Without you providing the actual links to texts you have used, we also have no way of examining what you have read for accuracy. That is why it is forum rules that you actually provide this information, which you still have not done after being asked to do so repeatedly.

Nowhere in relativity is it stated that things such as distances between two objects must remain the same regardless of the inertial frame. What is stated is that all inertial frames are equally valid as such and that the speed of light is invariant (ie, the same in all inertial frames). In fact, Galilean relativity (ie, classical mechanics) has a built in assumption that distances are the same in all inertial frames. This can be shown to be in direct conflict with the speed of light being the same in all inertial frames.
x-vision said:
In my mind, the distance between two bodies would be the same, irrespective of whether we look at it from the left-hand side or from the right-hand side, so to speak.
Nature does not care about what is in your mind, it works in a particular fashion that we can examine through observation and experimentation. It turns out that Nature does not work in the way that you imagine.

x-vision said:
In the reference frame of the rocket:
• The rocket is at rest and the star is moving towards the rocket.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the star will reach the rocket in 11.1 years.
• From the rocket's perspective, time is slowing down for the star, so only 4.8 years will have passed in the star's reference frame.
In the reference frame of the star:
• The star is at rest and the rocket is moving towards the star.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the rocket will reach the star in 11.1 years.
• From the star's perspective, time is slowing down for the rocket, so only 4.8 years will have passed in the rocket's reference frame.
The basic problem with your setup is the assumption that there's a meaning to "at time t(0)" that's shared between the two frames. There isn't (look up the Lorentz transforms and put ##t=t_0## into the time transform - you'll find that ##t'## isn't a single value, but a function of ##x##). So your two descriptions are descriptions of two different experiments.

As others have noted, you need a textbook. If you want a free one, former mentor @bcrowell has one downloadable from lightandmatter.com. I haven't read his SR one, but his GR one was interesting. Taylor and Wheeler's Spacetime Physics is pretty good in dead tree format, and the first chapter can be found on Taylor's website if you want a look.

x-vision and Dale
x-vision said:
I have calculated the 4.8 years interval using the time dilation formula
When starting to learn relativity I would strongly recommend not using the length contraction or time dilation formulas. You should stick exclusively with the Lorentz transform equations until you are more advanced. It is too easy to misuse the length contraction and time dilation formulas.

I will post your example worked out with the Lorentz transform later today.

Pencilvester and Ibix
Dale said:
When starting to learn relativity I would strongly recommend not using the length contraction or time dilation formulas. You should stick exclusively with the Lorentz transform equations until you are more advanced. It is too easy to misuse the length contraction and time dilation formulas.
Seconded. They're special cases of the full Lorentz transforms and it's very easy to apply them in more general contexts where they are invalid without realising what you are doing.

x-vision said:
In my mind, the distance between two bodies would be the same, irrespective of whether we look at it from the left-hand side or from the right-hand side, so to speak.
So, if we select either body as our stationary reference frame, we should see the same distance at time t(0)..
After that this distance will look different from the two reference frames, of course, but shouldn't it be the same at the starting point?

This contradicts the first postulate of SR that the speed of light is invariant (i.e. the same) across all inertial reference frames (IRFs). You either have

1) Newtonian Physics

Distances/lengths are invariant across all IRFs.
Elapsed times are invariant across all IRFs.
The speed of light is NOT invariant across all IRFs

Or, you have:

2) Special Relativity

The speed of light is invariant across all IRFs
Distances/lengths are NOT invariant across all IRFs
Elapsed times are NOT invariant across all IRFs.

Ibix
Ibix said:
Dale said:
When starting to learn relativity I would strongly recommend not using the length contraction or time dilation formulas. You should stick exclusively with the Lorentz transform equations until you are more advanced. It is too easy to misuse the length contraction and time dilation formulas.
Seconded. They're special cases of the full Lorentz transforms and it's very easy to apply them in more general contexts where they are invalid without realising what you are doing.
Thirded.

Pencilvester said:
Ibix said:
Dale said:
When starting to learn relativity I would strongly recommend not using the length contraction or time dilation formulas. You should stick exclusively with the Lorentz transform equations until you are more advanced. It is too easy to misuse the length contraction and time dilation formulas.

I will post your example worked out with the Lorentz transform later today.
Fourhted.

x-vision,

In your OP (original post), in the 1st half of your scenario description, you mentioned the star as stationary, and the rocket 10 ly distant moving inertially inbound toward the star at 0.9c. At that velocity, the gamma factor is indeed γ = 1/√(1-v²/c²) = 2.294. So when the rocket is at that precise range per the star (ie 10 ly), how far away is the star per the rocket (which holds itself as stationary, and the star moving)? In fact, the rocket then records the moving star to be at a contracted range of (10 ly) * (1/γ) = 10*(1/2.294) = 0.436 ly. That's not the 10 ly as specified in the 2nd half of your OP's scenario description. The star cannot be "both" 0.436 ly and 10 ly distant, per the rocket's POV as stationary. So your OP specifies 2 cases, and assumes those 2 cases are merely 2 different ways of looking at the very same relativistic scenario. But actually your OP specifies 2 completely different scenarios altogether, and while they are specified in a reciprocal fashion, they do not represent a single scenario viewed differently by each of 2 point-of-views (POVs) moving inertially relatively. When the star holds the rocket at 10 ly downrange, that rocket must then hold the star at 0.436 ly distant (not 10 ly).

Dale pointed out that the Lorentz Transformations should be learned first, rather than attacking relativity scenarios with only the time-dilation or length-contraction formulae. That is so true, assuming one wants an expedient learning curve. The Lorentz Transformations (LTs) map each point of one spacetime system to a unique point of another spacetime system moving relatively and inertially. The time-dilation and length-contraction equations are easily derivable from the LTs, and in doing so you will understand far better what they mean.

The LTs …

t’ = γ(t-vx/c²)
x’ = γ(x-vt)
y’ = y
z’ = z

γ = 1/√(1-v²/c²) ... gamma factor
1/γ = √(1-v²/c²) ... Lorents-Fitzgerald contraction factor

One more ... you should learn how Minkowski spacetime diagrams work. They are not hard at all. They'll usually save someone months of needless confusion, if learned early on. Likely, you can learn them in hours, or a few short days. Learning the meaning of the LTs in collective, hence the meaning of Special relativity, takes longer. But, spacetime figures will expedite that considerably.GrayGhost

x-vision
x-vision said:
Different articles on the Internet.

What articles? Please provide specific references. You have been asked for them repeatedly. If you do not provide them, this thread will be closed.

Also, as @Orodruin has already pointed out, "articles on the Internet" is not the right place to learn SR, or indeed any scientific field. You need to look at an actual textbook. Taylor & Wheeler's Spacetime Physics would be a good one to check out.

x-vision said:
In the reference frame of the rocket:
• The rocket is at rest and the star is moving towards the rocket.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the star will reach the rocket in 11.1 years.
• From the rocket's perspective, time is slowing down for the star, so only 4.8 years will have passed in the star's reference frame.

So I will use coordinates ##(t,x)## in the rocket frame and primed coordinates and variables ##(t',x')## in the star's frame. The rocket's worldline is ##r=(t,0)## and the star's worldline is ##s=(t,10-.9 t)##. The lines intersect when ##r=s## which is ##t=11.1##.

Lorentz transforming we get ##r'=(2.3 t, 2.1 t)## and ##s'=(20.6 + 0.44 t, 22.9)## in terms of t, so at ##t=0## the star's clock reads 20.6, and at ##t=11.1## when they meet the star's clock reads 25.5. So indeed the star's clock runs slow.

x-vision said:
In the reference frame of the star:
• The star is at rest and the rocket is moving towards the star.
• At time t(0), the distance between the rocket and the star is 10 light years.
• Since the distance between the two is 10 ly - and their relative velocity is 0.9c - the rocket will reach the star in 11.1 years.
• From the star's perspective, time is slowing down for the rocket, so only 4.8 years will have passed in the rocket's reference frame.

In the star's reference frame the worldlines (in terms of t') are ##r'=(t',0.9 t')## and ##s'=(t',22.9)##. Thus, at ##t'=0## the distance between the rocket and the star is 22.9. The rocket will reach the star in 25.5 years. From the star's perspective time is slow for the rocket so only 11.1 years will pass.

x-vision
Dale said:
So I will use coordinates ##(t,x)## in the rocket frame and primed coordinates and variables ##(t',x')## in the star's frame

Note that this is the opposite of what I recommended to the OP in post #3; you are assuming that his statement of the problem in the rocket frame is correct, and fixing the numbers for the star frame.

PeterDonis said:
Note that this is the opposite of what I recommended to the OP in post #3; you are assuming that his statement of the problem in the rocket frame is correct, and fixing the numbers for the star frame.
This got me thinking. If we reverse the assumption of which frame the problem is defined in then we will have the star at x’=0 in its rest frame and all of the numbers I wrote can be swapped between the star and the rocket. So what if neither were at x or x’=0?

It turns out that if the worldline of the rocket is ##r=(t,-6.96)## and the worldline of the star is ##s=(t,3.04-0.9t)## then when we transform into the star’s frame we get ##r’=(t’,-3.04+0.9t’)## and ##s’=(t’,6.96)##.

@x-vision it seems like there is indeed a way to set up the problem which meets your description in both frames. With this setup the distance is 10 ly both at t=0 in the rocket’s frame and at t’=0 in the star’s frame.

In both frames the other frame’s clock starts at 6.96 y. When they meet both clocks say 11.1 y. So according to each frame the other frame’s clock indeed advances slowly.

x-vision
Dale said:
it seems like there is indeed a way to set up the problem which meets your description in both frames. With this setup the distance is 10 ly both at t=0 in the rocket’s frame and at t’=0 in the star’s frame.

Note, however, that ##t = 0## in the rocket's frame and ##t' = 0## in the star's frame are not "the same time". They are two different spacelike lines, one passing through the event ##t = 0## on the rocket's worldline and one passing through the event ##t' = 0## on the star's worldline. These two lines do cross, but the point at which they cross is not on either worldline.

Dale said:
In both frames the other frame’s clock starts at 6.96 y.

I don't think this is quite correct. I get that at ##t = 0## in the rocket frame, the star is at event ##(0, 3.04)##, and ##t'## at that event is ##\gamma v x = 2.29 \cdot 0.9 \cdot 3.04 \approx 6.27##. To check this, since the star meets the rocket at ##t = 11.1## and the star's clock runs slow by the factor ##\gamma \approx 2.29##, we have ##11 . 1 / 2.29 \approx 4.84## years elapse on the star's clock, and ##4.84 + 6.27 \approx 11.1##.

x-vision and Dale
PeterDonis said:
Note, however, that t=0t=0t = 0 in the rocket's frame and t′=0t′=0t' = 0 in the star's frame are not "the same time". They are two different spacelike lines
Definitely, which is why I phrased it carefully. @x-vision should use a spacetime diagram to understand that important point.

PeterDonis said:
I don't think this is quite correct.
Oops, looks like I must have made an arithmetic error somewhere. I will check, but my time doesn’t give the right amount of time dilation so it must be wrong.

x-vision
Hi Dale. I see you guys were trying to make a single scenario that matched v-vision's OP, although I don't believe he had that in mind when he initially posted. It is a good idea though. I did draft the spacetime diagram here, however whereas you said "In both frames the other frame’s clock starts at 6.96 y" ... I get 6.268 y, not 6.96. Could you please reverify on your end? Thank you.

EDIT: Disregard, PeterDonis had just pointed out the same prior. Thanx

Best Regards,
GrayGhost

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Dale
x-vision said:
In my mind, the distance between two bodies would be the same, irrespective of whether we look at it from the left-hand side or from the right-hand side, so to speak.
As long as everything involved is motionless relative to every other thing in the scenario, then what you say is correct. Relativity deals with relative motion, though, so this example doesn't illustrate the equivalence of inertial reference frames.

x-vision
GrayGhost said:
whereas you said "In both frames the other frame’s clock starts at 6.96 y" ... I get 6.268 y, not 6.96.

See my post #23.

Here's a force-fitted spacetime diagram for x-vision's OP scenario, as he stated it. Please verify, thanks ...

GrayGhost

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x-vision
x-vision,

I reposted the spacetime diagram after you looked at it. I noticed I had omitted the negative signs on the (star's) x' values depicted there. They are corrected now.GrayGhost

x-vision
GrayGhost said:
I get 6.268 y, not 6.96.
Wow, this forum is awesome! This kind of fact checking really improves the accuracy and credibility of the forum.

x-vision said:
Given that there is no acceleration involved in this example, can we safely assume that the two reference frames are fully symmetrical?
Yes, we can. The usual Lorentz contraction scenarios are not symmetrically.

I extend your scenario by an additional object and call it "Earth". The Earth is 10 LJ away from the star.

In the reference frame of the star:

The rocket is 10 LJ away from the star, when the rocket passes the Earth.

In the reference frame of the rocket:

The star is 10 LJ away from the rocket, before the Earth passes the rocket.

x-vision,

While my prior spacetime diagram was correct in its depicted coordinates, it was technically incorrect from a standpoint of "proper convention" for spacetime diagrams. Proper convention, is to present the system which holds the velocity as positive (worldines slant from lower-left to upper-right) as unprimed, which is usually depicted as stationary (verticle). From the point of view (POV) of the other system, that system records a negative velocity (worldines slant from lower-right to upper-left). For that system, the "Inverse LTs" apply.

The LTs ...

t’ = γ(t-vx/c²)
x’ = γ(x-vt)
y’ = y
z’ = z

γ = 1/√(1-v²/c²)

Inverse LTs ...

t = γ(t’+vx’/c²)
x = γ(x’+vt’)
y = y’
z = z’

γ = 1/√(1-v²/c²)

My next figure will have the STAR frame depicted as stationary, unprimed, with black solid worldline and black dashed (horizontal) lines-of-simultaneity. The ROCKET frame will be depicted as moving, primed, and with blue worldine and blue dashed (slanted) lines-of-simultaneity.

The inverse LTs are derived from the usual standard LTs, by merely plugging in -v for v into the LTs. The only reason this is done, is because many trying to learn SR often confuse the systems. With the above convention, you no longer have to be concerned with the velocity's polarity when using the transforms, although you do need to remember which system to present as unprimed. The only exception is the light path, which is always at 45 degrees in either direction, and which also bisects the time and spatial axes of all depicted systems, moving or not.

If you have any questions regarding the spacetime figures, feel free to ask.GrayGhost

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x-vision,

Wrt my prior post #28, your OP scenario's spacetime diagram redone with with proper convention ...

GrayGhost

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x-vision,

BTW, for your scenario the LTs apply, but only after some modification. The LTs were derived for spacetime systems that have a common origin. IOW x,y,z,t = x',y',z',t' = 0,0,0,0. Their spatial origins momentarily coincide at one point in time, when both their clocks read zero. In your scenario, as you had defined it, this is not the case. Your 2 systems momentraily coincide when both clocks read 10ly/v = 11.11 y. As such, I had to modify the LTs for the 11.11 y time offset. Here are the equations I used ...

So, I used the modified LTs above.

Try plugging in star frame coordinates to determine corresponding rocket frame coordinates, as per my spacetime figure. Also, try plugging in rocket frame coordinates to determine corresponding star frame coordinates. Play with it a bit, and it gets pretty easy.

You might try your same scenario assuming they meet up when both their clocks read 0, versus 11.11. Then you just use the standard LTs and Inverse LTs, and no longer need the Modified LTs above. Feel free to ask questions, as I'd be happy to assist. Hope that helps.GrayGhost

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Dale said:
GrayGhost said:
I get 6.268 y, not 6.96.
Wow, this forum is awesome! This kind of fact checking really improves the accuracy and credibility of the forum.

OK Dale. You're right, I shall try to do better in that respect. Here's what I had for that ...

Best Regards,
GrayGhost

Dale
<h2>1. What is reference frame symmetry in Special Relativity?</h2><p>Reference frame symmetry in Special Relativity refers to the idea that the laws of physics should be the same for all inertial reference frames. In other words, the physical laws should appear the same to all observers who are moving at a constant velocity relative to each other.</p><h2>2. Why is reference frame symmetry important in Special Relativity?</h2><p>Reference frame symmetry is important in Special Relativity because it is a fundamental principle that helps to explain how the laws of physics work in different reference frames. It also helps to reconcile the apparent differences in measurements made by observers in different reference frames.</p><h2>3. How does reference frame symmetry affect time and space in Special Relativity?</h2><p>Reference frame symmetry in Special Relativity leads to the concepts of time dilation and length contraction. This means that time and space are perceived differently by observers in different reference frames, depending on their relative velocities.</p><h2>4. Can reference frame symmetry be violated in Special Relativity?</h2><p>No, reference frame symmetry is a fundamental principle in Special Relativity and has been confirmed by numerous experiments. It is a cornerstone of the theory and is considered to be a universal law of nature.</p><h2>5. How does reference frame symmetry relate to the theory of relativity?</h2><p>Reference frame symmetry is a key concept in the theory of relativity. The theory of Special Relativity is based on the principle of reference frame symmetry, while the theory of General Relativity extends this principle to include non-inertial reference frames. Both theories rely on the idea that the laws of physics should be the same for all observers, regardless of their relative motion.</p>

## 1. What is reference frame symmetry in Special Relativity?

Reference frame symmetry in Special Relativity refers to the idea that the laws of physics should be the same for all inertial reference frames. In other words, the physical laws should appear the same to all observers who are moving at a constant velocity relative to each other.

## 2. Why is reference frame symmetry important in Special Relativity?

Reference frame symmetry is important in Special Relativity because it is a fundamental principle that helps to explain how the laws of physics work in different reference frames. It also helps to reconcile the apparent differences in measurements made by observers in different reference frames.

## 3. How does reference frame symmetry affect time and space in Special Relativity?

Reference frame symmetry in Special Relativity leads to the concepts of time dilation and length contraction. This means that time and space are perceived differently by observers in different reference frames, depending on their relative velocities.

## 4. Can reference frame symmetry be violated in Special Relativity?

No, reference frame symmetry is a fundamental principle in Special Relativity and has been confirmed by numerous experiments. It is a cornerstone of the theory and is considered to be a universal law of nature.

## 5. How does reference frame symmetry relate to the theory of relativity?

Reference frame symmetry is a key concept in the theory of relativity. The theory of Special Relativity is based on the principle of reference frame symmetry, while the theory of General Relativity extends this principle to include non-inertial reference frames. Both theories rely on the idea that the laws of physics should be the same for all observers, regardless of their relative motion.

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