Why is invariant interval invariant?

[kof9595995]

The invariant interval is defined to be
\Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}
and despite which inertial frame we are in, \Delta s for two particular events would be the same.
If I use Lorentz transformation, this can be proved easily. But is there any more "intuitive" way to verify the invariance? Like when (x,y,z,t) describes propagation of light, it'll be trivially true,e.g. \Delta s=0 because of the principle of constancy of light velocity.
But what about other cases, when two events don't lie on the same light cone? Of course \Delta s is not 0 but still remains invariant, but how to convince myself it's true without doing the arithmetic manipulation of Lorentz transformation?

 

[DrGreg]

It's not exactly intuitive, but you might find this post relevant.

 

[Bob_for_short]

The invariant interval is defined to be
\Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}
and despite which inertial frame we are in, \Delta s for two particular events would be the same.
If I use Lorentz transformation, this can be proved easily. But is there any more "intuitive" way to verify the invariance? Like when (x,y,z,t) describes propagation of light, it'll be trivially true,e.g. \Delta s=0 because of the principle of constancy of light velocity.
But what about other cases, when two events don't lie on the same light cone? Of course \Delta s is not 0 but still remains invariant, but how to convince myself it's true without doing the arithmetic manipulation of Lorentz transformation?

It's easy. A space-like interval is just a length, distance in the frame of reference where delta_t=0. I.e., it is a tangible physical thing. In other RFs you can obtain its value too if you make the neccesary combination of delta_t' and delta_x'.
The same is valid for a time-like interval. It is just one hour. It is normal that this result can be obtained in other RFs - they also have means to recalculate their observations into your system. It would be highly frustrating not to be able to have a rule to get such an information exchange between physical RFs.

 

[Meir Achuz]

Think of the Lorentz T as a rotation in Minkowski space. Then the length \Delta S doesn't change.

 

[Dale]

Think of the Lorentz T as a rotation in Minkowski space. Then the length \Delta S doesn't change.

That is my preferred visualization also. Then it comes down to realizing that in Minkowski space "lengths" are defined by hyperboloids instead of circles. That realization also makes it clear why you cannot rotate a timelike interval into a spacelike interval and also why a future-directed timelike interval cannot be rotated into a past-directed timelike interval.

 

[kof9595995]

Thank you guys, but I think the rotation point of view is also derived from LT. Actually I was asking if we can skip the LT and verify the invariance just by two postulates.

 

[Dale]

Thank you guys, but I think the rotation point of view is also derived from LT. Actually I was asking if we can skip the LT and verify the invariance just by two postulates.

No, that is certainly not possible. The term "invariant" means that it does not change under the Lorentz transform. So it is certainly not possible to say that something does not change under a transform without defining the transform (other than by tautology).

 

[Bob_for_short]

We can look at it without LT. Let us see a stick moving fast along us. It bears a label "WalMart". We can make two measurements of its "length". One measurement will give a contracted length and another can give the length in our RF if the stick were still (its proper length = interval). The latter case needs special experiment but no LT. Just two values: delta_x and delta_t. The numerical value of the interval answers the question about the stick length at rest in our RF if we would buy it at WalMart. So you may think of invariance of the interval as of invariance of the stick length from WalMart however fast you move after buying it.

 

[Meir Achuz]

Thank you guys, but I think the rotation point of view is also derived from LT. Actually I was asking if we can skip the LT and verify the invariance just by two postulates.

You have it backwards. The physical principle is that going to a moving coordinate system corresponds to a generalized rotation in space-time. The LT can be derived from that.

 

[kof9595995]

No, that is certainly not possible. The term "invariant" means that it does not change under the Lorentz transform. So it is certainly not possible to say that something does not change under a transform without defining the transform (other than by tautology).

But still, is it possible to skip the explicit formula of LT, and verify the invariance.
I mean, certainly invariance only makes sense under some transformation. But now we know the real transformation is not Galilean, but something can make the speed of light constant. So I want to ask, is it possible to verify the invariance under this "real transformation" pretending we don't know such thing as Lorentz formula?

 

[kof9595995]

We can look at it without LT. Let us see a stick moving fast along us. It bears a label "WalMart". We can make two measurements of its "length". One measurement will give a contracted length and another can give the length in our RF if the stick were still (its proper length = interval). The latter case needs special experiment but no LT. Just two values: delta_x and delta_t. The numerical value of the interval answers the question about the stick length at rest in our RF if we would buy it at WalMart. So you may think of invariance of the interval as of invariance of the stick length from WalMart however fast you move after buying it.

That is a good way to think of it, space-like region gives the proper length and time-like gives proper time, but still I can't see why if I don't apply the Lorentz T explicitly

 

[Dale]

But still, is it possible to skip the explicit formula of LT, and verify the invariance.

Not that I know of (except tautologically).

 

[Bob_for_short]

That is a good way to think of it, space-like region gives the proper length and time-like gives proper time, but still I can't see why if I don't apply the Lorentz T explicitly

We can understand it from an ordinary relativity and ambiguity of measurements.

For example, if a direct measurement of the proper length can be done by arranging a ruler along the stick, in a distant but still RF you with your ruler will obtain a different value because of the perspective law. You have to transform your data with help of a certain rule to get the proper length (which is, by the way, is visible on the first ruler).

Similarly with the interval - you should not think of its invariance at all but of the proper length solely. Think of proper length as of certainty, not as of invariance.

delta_x and delta_t (your data) squared in the interval are not the LT. The interval formula is just a rule of transforming your measurement data to learn the proper length: there is no relative velocity involved in the interval!

The only "postulate" needed to assure you is that it is always possible to do.

 

[kof9595995]

Hi Bob, I'm totally lost by your post 13, can you explain a bit more? And what do you mean by " a distant but still RF"

 

[Bob_for_short]

Hi Bob, I'm totally lost by your post 13, can you explain a bit more? And what do you mean by " a distant but still RF"

A distant but still RF is a reference frame resting at some distance from the stick with its own ruler attached. No relative velocity. You just use your eye, your ruler and your hand (the stick is out of reach). So it looks shorter, isn't it? Knowing the distances - from the stick to you, the ruler readings, and your hand length, you calculate the proper stick length.

 

[Dale]

In SR reference frames have infinite spatial extent, so there is no such thing as a reference frame being located some distance from a specific object. I think you simply mean a reference frame where the object is at rest at some location other than the origin.

Having said that can you show how your idea of measuring lengths using similar triangles results in the formula for the invariant interval without using the Lorentz transform? It is certainly not clear to me that it is possible to derive the interval formula that way.

 

[Bob_for_short]

...can you show how your idea of measuring lengths using similar triangles results in the formula for the invariant interval without using the Lorentz transform? It is certainly not clear to me that it is possible to derive the interval formula that way.

It does not result in the formula for the interval. I never said that! I said usual (daily) relativity, not special relativity. I meant to say that measuring something (in any physical situation) in a RF different from the proper RF needs recalculation or transformation of the data to get the proper values. The proper values are certain. Any proper value is unique. Interval is just a proper value of time or distance. It is invariant because it is unique, certain, tangible.

 

[Dale]

I don't think that answers the OP's question.

 

[DrGreg]

But still, is it possible to skip the explicit formula of LT, and verify the invariance.

Not that I know of (except tautologically).

I disagree...

...you might find this post relevant.

 

[lalbatros]

kof9595995,

You are right that light rays in inertial frames obey the differential equation ds²=0.
Therefore ds²=0 in one inertial frame R should imply ds'²=0 in any other inertial frame R'.
This is only possible if there is this kind of transformation of the element from frame R to frame R':

[1] . . . ds'² = a(V) ds²

where V is the relative velocity of R' with respect to R.
More complicated relations should be excluded based on the uniformity of space-time.

Consider now a frame R" with the opposite relative velocity with respect to R.
We should have:

[2] . . . ds"² = a(-V) ds² .

We expect to have ds'² = ds"² because of the symmetry and therefore

[3] . . . a(V) = a(-V)

In addition, the transformation from R' to R should be based on the opposite velocity (-V) as compared to [1]:

[4] . . . ds² = a(-V) ds²

Therefore by combining [1] and [4]:

[5] . . . a(V)a(-V) = 1

Combining [3] and [5] leads to a(V) = 1.

The relation [1] defines a group.
The symmetry conditions lead to a very trivial group!

Michel

 

[Bob_for_short]

...Combining [3] and [5] leads to a(V) = 1.

If ds²=0 and ds'²=0, then a(V) may be chosen to be 1.1 too.

 

[pervect]

"Why" questions can be asked recursively. At some point, you have to stop asking them. Kids learn that early - they can always ask "but why..." to the answer to any why question.

Why questions aren't totally hopeless though - sometimes you can explain the "why" of complex phenomenon in terms of simpler principles.

So I view it as useful to answer "why" questions up to some level which is the most fundamental.

In my opinion, the invariance of the Lorentz interval is one of the most fundamental principles of relativity - I doubt there is anything that's simpler that's more fundamental to the theory.

 

[Bob_for_short]

...In my opinion, the invariance of the Lorentz interval is one of the most fundamental principles of relativity - I doubt there is anything that's simpler that's more fundamental to the theory.

It was Henri Poincaré who first introduced and proved invariance of the interval as the invariance of a distance in a pseudo-euclidean space-time under four-rotations (LT).

Any invariance (not only SR's ones) means a possibility to learn proper properties from measurements made in different reference frames. It is a banality in Physics.

 

[Dale]

I disagree...you might find this post relevant.

Very good proof! I think that answers the OP's question nicely, and my previous posts to the contrary are clearly incorrect.

 

[lalbatros]

If ds²=0 and ds'²=0, then a(V) may be chosen to be 1.1 too.

No.
I explained the additional symmetry requirements.
Relation [1] is about the homogeneity of space.
Relation [3] is related to the isotropy of space.
Relation [5] requires that forth and back transformation leave the ds² invariant.
Relation [5] actually requires that the ds² transformation makes a group.

Combining these relations:

[3] . . . a(V) = a(-V)
[5] . . . a(V)a(-V) = 1

leads to a(V)a(V) = 1.

From this you have only two choices: a(V)=1 or a(V)=-1.
Only the first choice makes sense.

Your quick pick that a(V)=1.1 fails on the group requirement [5].

You can see similar arguments applied to derive the LT in the Einstein's historical paper, including the group requirement.
(see §3 in http://www.fourmilab.ch/etexts/einstein/specrel/www/)

 

[Bob_for_short]

It makes sense to speak of a(V) if ds ≠ 0. And you started from ds = 0.

Concerning the group character of LT, it was first written in H. Poincaré seminal article.
He required the mechanical equations to be also invariant under LT, i.e., introduced ds ≠ 0.

See http://www.univ-nancy2.fr/poincare/bhp/hp1906rpen.xml

 

[lalbatros]

It makes sense to speak of a(V) if ds ≠ 0. And you started from ds = 0.

kof9595995 was asking about why ds² should be invariant.
Therefore I was implicitely discussing the transformation from ds² to ds'² in the general case.
I skipped a few obvious steps to go directly to the consequence of the equivalence of ds²=0 and ds'²=0.

To be more explicit, the general form of the transformation could be this:

ds'² = f(x,y,z,t,x',y',z',t',Vx,Vy,Vz,ds²)

Obvious arguments (same as those listed in Einstein 1905) lead to the simplification:

ds'² = f(V,ds²)

Developping this in series, assuming ds² is small enough, leads to

[6] . . . ds'² = b(V) + a(V) ds²

This relation [6] should be valid for any interval, even along light rays.
Since we should have ds²=0 => ds'²=0, we can conclude that:

b(V) = 0

And this leads us to:

[1] . . . ds'² = a(V) ds²

This is were I started my https://www.physicsforums.com/showpost.php?p=2440322&postcount=20".

But https://www.physicsforums.com/showpost.php?p=2437046&postcount=3", this was indeed an easy question.

 

[kof9595995]

It's not exactly intuitive, but you might find this post relevant.

Hi DrGreg, you proved dedr=de'dr', and denote it as ds^2. But I don't think it's good enough, you've shown according to first postulate, ds^2=dedr is invariant, but you didn't show further that \Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}
So the invariance of dedr is not so satisfactory because I can also argue that dedr=dt^2, which bring us back to Galilean transformation

 

[DrGreg]

Hi DrGreg, you proved dedr=de'dr', and denote it as ds^2. But I don't think it's good enough, you've shown according to first postulate, ds^2=dedr is invariant, but you didn't show further that \Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}
So the invariance of dedr is not so satisfactory because I can also argue that dedr=dt^2, which bring us back to Galilean transformation

Carry on reading my post. I found e and r in terms of x and t and said "Substitute (4) and (5) into (3) to get the standard equation for the Lorentz interval." I didn't give all the details (an exercise for the reader!) and I ignored y and z, but once you've got the result in 2D spacetime, it's not hard to generalise to 4D spacetime.

If you can't work out the missing details, ask and I'll help.

Just in case you're confused, in my version I'm using the convention that

ds^2 = dt^2 - \frac{dx^2}{c^2}​

If you want consistency with your version I'd have to redefine ds^2 = -c^2\,de\,dr.

P.S. the link in my original post is now dead. Site moved to http://autotheist.synthasite.com/bondi1.php

 

[kof9595995]

Sorry I did not read your post carefully enough. Your post do seem to be the nicest answer to my question so far, though also as you said, "not exactly intuitive".
Thanks