Transformation Vs. Physical Law

[Austin0]

You are right that the gamma factor was described first, and that its discovery set us on the path that led to the discovery of the four-vector formulation of SR, the Minkowski metric, and eventually GR. But that's the history, not the logical structure of the theory that we see when we've made it to the other end of that path.

A perhaps less controversial example: the discovery that dropped objects accelerate towards the surface of the Earth at 10 m/sec^2 preceded Newton's principle of gravitation. But it would be absurd to argue from this history that Newtonian gravity and and G requires 10 m/sec^2 and not the other way around.

i think maybe your analogy is a little distorted. The gamma function is more analogous to the G of gravity as being a fundamental aspect of physics than it is to any specific velocity or value. Newton inferred gravity from observation while Lorentz derived gamma from the Maxwell math but they were both discoveries of fundamental principles

 

[Samshorn]

$d\tau^2=dt^2-dr^2-r^2d\theta^2-dz^2$ can be written as $d\tau^2=dt^2 / \gamma^2$...

No it can't, at least, not without doing a coordinate transform into a standard rectilinear coordinate system.

I can't imagine why you think that. Those two equations are mathematically equivalent, just two different ways of writing the same thing. Just factor dt^2 out of the right hand side of the first equation. The remaining factor is (1 - v^2), which is 1/gamma^2.

There is never any need to use t as the parameter.

Not true. For example, when trying to determine the elapsed t, we obviously need to use the t coordinate. In your example you just split up the computation into two separate calculations, one for tau and one for t, and declined to note the ratio, but ordinarily the ratio is what we want. In other words, knowing the half-life in the rest frame of the particle, we want to know the half-life in terms of the t of some other coordinate system.

Non inertial reference frames do not, in general, have factors of gamma.

It's always possible to express the ratios of times in terms of gamma factors. I think the first step toward understanding this would be to understand why those two expressions above are mathematically equivalent.

 

[Jorriss]

But instead we should have a physical law explaining these differences, which then can be validly transformed for any other inertial observing frame using Lorentz transformation.

There is physical law behind it. It's that the speed of light is the same in all reference frames. If the speed of light is the same in all references then a moving clock will appear to run slow which can be quantitatively described by lorentz transformations.

 

[harrylin]

I meant where exactly in my calculation. I agree that gamma is part of the Lorentz transform, but I never did a Lorentz transform in my calculation. I also agree that gamma can be derived from the metric, but that derivation wasn't part of my calculation. So it is quite a stretch, IMO, to say that gamma is in there, even implicitly.[..]

The "metric" that you used for your calculation is effectively the cited "quadratic form" that is equivalent to the Lorentz group (except that you used polar coordinates). The same calculation can be done using gamma explicitly.

[..] Don't you consider the invariant interval equation a Lorentz transformation?? [..]

Yes, as I cited, the invariant interval equation was introduced as an alternative (but equivalent) description of the Lorentz group.

 

[harrylin]

I already agree to that fact, that every calculation involving factor gamma is not a transform. Instead this was the one of the points in my original post.

The question was if it does not come from the transformations then where does this gamma factor come from.

But I think, it's pointless, to discuss it here. [..]

OK. Well, obviously it is part of laws of nature. And the same question can be asked about all laws of nature. So, perhaps that becomes too philosophical indeed!

However, partial answers exist. Special relativity assumes that everything behaves like electromagnetism - and that is only a small stretch from knowing that matter is governed by electromagnetic bonds. From such considerations one can build special relativity "bottom-up" (and in fact this is just how the early development proceeded), for example by analyzing how a light clock would behave in motion.
- http://en.wikipedia.org/wiki/Time_d...nce_of_time_dilation_due_to_relative_velocity

 

[zonde]

There is physical law behind it. It's that the speed of light is the same in all reference frames. If the speed of light is the same in all references then a moving clock will appear to run slow which can be quantitatively described by lorentz transformations.

This is not true. Speed of light being the same in all admissible reference frames is a postulate (not a physical law) and it effectively defines "all admissible reference frames".

Physical laws are defined within single reference frame not across many reference frames. So if you try to define speed of light within single reference frame you get physical constant with particular value. Then in a sense we can say that if we treat speed of light as physical constant (kind of trivial physical law) then the statement that "speed of light is the same in all admissible reference frames" becomes redundant as it is already implied by principle of relativity.

 

[A.T.]

Then in a sense we can say that if we treat speed of light as physical constant (kind of trivial physical law) then the statement that "speed of light is the same in all admissible reference frames" becomes redundant as it is already implied by principle of relativity.

No, the 2nd postulate in not redundant and is not implied by the 1st postulate. The ballistic light theory satisfies the 1st but not the 2nd postulate, so there is no implication 1 -> 2.

The 2nd postulate is not just that speed of light is the same in all inertial frames, but also that it is independent of the source velocity in all inertial frames. So the 2nd postulate effectively says that the same light beam, has the same speed across all inertial reference frames. And as you correctly stated "physical laws" in the sense of the 1st postulate are defined within single reference frame, so you cannot lump the 2nd postulate as a "physical law" in there.

 

[Dale]

Hi, DaleSpam. Could you show how this is established?

The place that I learned it was primarily chapter 2 of Sean M. Carroll's lecture notes on GR:
http://arxiv.org/abs/gr-qc/9712019

I list here the fundamental geometric objects of GR, not in the order of presentation in Carroll, but in order of generality:
Topological spaces (continuity)
Manifolds (continuity, dimensionality)
Metric (continuity, dimensionality, distance and angles)
Coordinate chart (continuity, dimensionality, distance and angles, labelling)
Coordinate transforms (continuity, dimensionality, distance and angles, labelling, correspondence)

At each level, everything from all the previous levels is included and something new is added. So, for example, all manifolds are topological spaces, but not all topological spaces are manifolds, making topological spaces more fundamental/general than manifolds.

 

[harrylin]

No, the 2nd postulate in not redundant and is not implied by the 1st postulate. The ballistic light theory satisfies the 1st but not the 2nd postulate, so there is no implication 1 -> 2.

The 2nd postulate is not just that speed of light is the same in all inertial frames, but also that it is independent of the source velocity in all inertial frames. So the 2nd postulate effectively says that the same light beam, has the same speed across all inertial reference frames. And as you correctly stated "physical laws" in the sense of the 1st postulate are defined within single reference frame, so you cannot lump the 2nd postulate as a "physical law" in there.

As that's a bit off-topic, just this for the record: That is a myth[2].
The second postulate is a physical law which adheres to Maxwell's wave model and rejects ballistic emission theory; it is defined for a single reference frame. See:
[1] http://www.fourmilab.ch/etexts/einstein/specrel/www/
[2] http://www.uio.no/studier/emner/matnat/fys/FYS-MEK1110/v06/MythsSpecRelativAJP193.pdf

 

[Dale]

The metric did not appear out of thin air and how could it?

In GR, often times metrics do appear "out of thin air". I.e. it is often easier to find a solution to the EFE by starting with the metric and then solving for the distribution of mass and energy required to make that metric.

You didn't answer the actual question A

The actual question was based on a flawed analogy. So I tried to de-analogize it and answer the underlying question, which I though was a valid question despite the bad analogy.

As far as where it might be lurking, I suspect that harrylin might be on the right track.
the Pythagorean operation returns the value of a line interval in Minkowski geometric space.
... So as the pythagorean operation does perform this transformation, it is in effect a geometric gamma function.

First, I disagree, and second, I never used the Pythagorean theorem nor the Minkowski metric. You are not getting any closer to showing where I used gamma in my calculations.

More importantly; as you agree that the result contains the gamma factor and I am sure you would agree that this factor was not implicit in the raw coordinate values, the question becomes ,where else could it possibly be lurking??

It lurks in differeniating the line element wrt coordinate time in Minkowski coordinates. I neither used Minkowski coordinates nor did I differentiate wrt coordinate time. You cannot get from what I did to gamma without transforming to Minkowski coordinates and differentiating wrt coordinate time, neither of which are required for calculating the decay of the muons.

Well as it seems clear it is a transformation

Nonsense. A transformation requires at least two different coordinate systems and a mapping between them. The metric doesn't even require coordinate systems, let alone a pair of them. So claiming that it a transformation is wrong, and claiming that it is clearly a transformation is absurd.

 

[Dale]

Those two equations are mathematically equivalent

Yes. But mathematical equivalency does not imply that calculating gamma was necessary in order to apply a physical law to predict the outcome of a given scenario. For example, you can do Newtonian mechanics without calculating Lagrangians, and you can do Lagrangian mechanics withoug calculating Newtonian forces. The fact that they are equivalent does not mean that one is required when doing the other, although you can transform between the two whenever desired.

I said from the beginning that it was possible to derive gamma from what I did, I was merely pointing out that it was not an essential part of the calculation. In fact, the mathematical equivalency strengthens my argument, since it means that in a potentially large class of problems it is possible to avoid using gamma. Similarly to the large class of problems where it is possible to avoid using forces by using Lagrangians.

For example, when trying to determine the elapsed t, we obviously need to use the t coordinate.

But I never tried to determine the elapsed t.

It's always possible to express the ratios of times in terms of gamma factors. I think the first step toward understanding this would be to understand why those two expressions above are mathematically equivalent.

I am skeptical that it is always possible (e.g. curved spacetimes or null coordinates), but even if it is possible, it is not necessary. My point has not been that it is not possible to determine gamma, but that it is not necessary to do so in applying physical laws.

 

[Dale]

The same calculation can be done using gamma explicitly.

Yes, but that doesn't mean that I did it using gamma implicitly.

 

[harrylin]

Yes, but that doesn't mean that I did it using gamma implicitly.

Perhaps "implicit" is too strong a word and I should have written "contained" or "equivalent"? Anyway, my point was that tricks to avoid gamma surely didn't help universal and I think that everything has been sufficiently clarified by now.

 

[Dale]

To everyone:

It is very easy to get gamma from the Minkowski coordinates (which I didn't use) by differentiating the line element wrt coordinate time (which I didn't do):$$c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$$$$\frac{d\tau^2}{dt^2} = 1 - \frac{1}{c^2}\left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right)$$$$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}=\gamma^{-1}$$

That is clear, but that is not the point I was making to the OP. The point is that you are not required to do either of those things (use Minkowski coordinates or parameterize wrt coordinate time) in order to use the physical law as cited. I.e. calculating and using gamma is possible, but not necessary.

The OP seems to think that any use of gamma implies a coordinate transform, to which I disagree, but he also though that use of gamma was necessary for the use of the decay law. I was refuting the latter claim.

 

[Dale]

Perhaps "implicit" is too strong a word and I should have written "contained" or "equivalent"?

I agree to "equivalent".

 

[Samshorn]

It is very easy to get gamma from the Minkowski coordinates (which I didn't use) by differentiating the line element wrt coordinate time (which I didn't do).

I think there are several things wrong with what you've been saying. First, your "lab" analysis made no sense (as I'll explain). Second, your cyclotron analysis was simply integrating gamma (which I've already explained, and will try to clarify). Third, none of this has anything to do with the choice of Cartesian or polar space coordinates. Fourth, differentiation isn't necessary to relate gamma to the line element, since it is simple algebra.

Okay, let's take these one at a time. First, your lab analysis was totally nuts, because you defined "a" as the number of laps around the cyclotron and T as the time (implicitly measured in the lab frame coordinate time t) to make one lap. But muons at rest in the lab are making no laps at all, so a=0 and T is infinite. Your result is "aT", which makes no sense for muons at rest. To clean up this mess, you need to dump "a" and T, and all you are left with is dtau = dt for a particle at rest relative to the inertial coordinates with time coordinate t, since the space coordinates are constant.

Second, your cyclotron analysis simply amounted to integrating (1/gamma)dt, disguised and made needlessly convoluted by replacing the coordinate time t with aT and integrating over "a" instead of over t. You defined t = aT where T is a constant for a given v defined by v = 2pi R/T, so "a" is just a re-scaled expression for the coordinate time t. Naturally since you are just integrating (1/gamma)dt you arrive at the result tau = t/gamma. And then you claim that neither the coordinate time t nor gamma are involved in your calculation!

Third, your comments about polar coordinates versus Cartesian coordinates are puzzling, because it makes no difference what space coordinates we use. The spatial part of the metric can simply be abbreviated as dS, which is the appropriate function of the coordinate differentials. And v is dS/dt in terms of this coordinate system, regardless of whether we use Cartesian or polar or any other system of space coordinates.

Fourth, when you say we can get gamma by differentiating the metric with respect to coordinate time, you overlook the fact that the metric is already a differential expression, so it is simple algebra to divide by dt. No differentiation is required. That's why the metric and the equation dtau = dt/gamma are algebraically equivalent, not requiring any calculus or differentiation to relate them.

The point is that you are not required to do either of those things (use Minkowski coordinates or parameterize wrt coordinate time) in order to use the physical law as cited.

Well, it's true that we can use whatever coordinate systems we like, but they must either be inertial coordinates or else the expression for the line element must be modified to compensate for the non-inertial coordinates, precisely to the extent that they are non-inertial, so in effect we are still using inertial coordinates. And of course the use of polar space coordinates is trivial, since we still have dtau^2 = dt^2 - dS^2 where t is a standard inertial time coordinate and dS is the space differential. Also, when you say we can parameterize by something other than t, well, if we define a re-scaled version of t, such as a = t/T for some constant T, then sure, but to argue that we are no longer using t is rather silly. It's just a re-scaled t, i.e., it is a choice of units, nothing more.

The OP seems to think that any use of gamma implies a coordinate transform, to which I disagree...

I agree that the OP was totally wrong about that (see post #52).

He also though that use of gamma was necessary for the use of the decay law. I was refuting the latter claim.

I don't think your refutation holds water, for the reasons explained above. Mind you, we can certainly contrive to avoid writing the greek symbol "gamma", merely by writing out the definition of gamma in full, which basically is what the metric line element represents. (Likewise we can avoid writing "v" by writing dS/dt, but would we really claim we have avoided using v?) But I don't think the OP's fundamental error is in thinking that the results of special relativity are represented by the gamma factor. The gamma factor actually does encode the essential non-positive-definite signature of the Minkowski metric, from which the unique effects of special relativity arise. So although associating everything with "the gamma factor" may be a somewhat dim-witted way of looking at things, it isn't exactly wrong.

The OP's fundamental problem, as he clarified in his "farewell cruel world" post is that he says he wants to know "where this factor comes from", and yet he in unable to articulate what he means by this question. He began by saying he would be satisfied if someone could give him the physical law, not involving a transformation, but then when you provided that law he shifted his ground, and began asking where that law "comes from".

Obviously that question is so vague as to be meaningless, and all efforts to get him to clarify his meaning are doomed to fail, basically becuase he doesn't have any clue what he means, because he has never subjected his own beliefs to any kind of rational scrutiny. My guess is that the only answer about where something "comes from" that would satisfy him is an explanation that conforms to his personal pre-conceptions, prejudices, and misconceptions, none of which he ever intends to give up. Anything else he will simply reject as not satisfactory. Still, it's sometimes of interest to engage someone like that in conversation, if only for the light it sheds on some pathological aspects of human psychology.

 

[Mentz114]

$$c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$$$$\frac{d\tau^2}{dt^2} = 1 - \frac{1}{c^2}\left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right)$$$$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}=\gamma^{-1}$$

Most math literate people work this out pretty quickly - but it should be taught emphatically and early in all SR courses. The metric defines gamma, and in my view comes before gamma logically. It's the metric that welds space and time together, all the other stuff follows.

 

[Dale]

Most math literate people work this out pretty quickly - but it should be taught emphatically and early in all SR courses. The metric defines gamma, and in my view comes before gamma logically. It's the metric that welds space and time together, all the other stuff follows.

I agree.

 

[Dale]

Hi Samshorn, your post is way to long to respond to in detail, so I will only pick a few things. Overall, I wanted to show that you could use the law to predict the outcome of the specified experiment without using gamma. I used the law for the specified experiment. I didn't use gamma. So IMO I did what I set out to do. I agree that you can start from what I did and do things differently and get gamma, but that doesn't change the fact that I did exactly what I claimed.

First, your "lab" analysis made no sense (as I'll explain). Second, your cyclotron analysis was simply integrating gamma (which I've already explained, and will try to clarify). Third, none of this has anything to do with the choice of Cartesian or polar space coordinates. Fourth, differentiation isn't necessary to relate gamma to the line element, since it is simple algebra.

I agree with your fourth point, but the lab analysis was perfectly fine (a is a completely valid parameterization of the worldline in both cases even if you think it is weird), I didn't integrate gamma (if I had then I would have gotten the wrong number since the limits of integration were 0 to 1 instead of 0 to T), and the choice of coordinates is essential since gamma is a feature of a specific set of coordinates (Minkowski) on a specific spacetime (flat).

Third, your comments about polar coordinates versus Cartesian coordinates are just weird. You can't really believe it makes any difference. The spatial part of the metric can simply be abbreviated as dS, which is the appropriate function of the coordinate differentials. And v is dS/dt in terms of this coordinate system, regardless of whether we use Cartesian or polar or any other system of space coordinates.

I see your point here although your first two sentences are a little rude, but even collecting the spatial terms into dS the fact remains that dS/dt was not a part of my analysis. As I have said many times already, you certainly can branch off from what I did and get a gamma, but as I showed it is not necessary.

The gamma factor actually does encode the essential non-positive-definite signature of the Minkowski metric, from which the unique effects of special relativity arise.

I don't know what you mean here. The signature of the metric is a different thing from gamma. Consider Schwarzschild coordinates, it has the same signature, but doesn't have gamma.

The OP's fundamental problem, as he clarified in his "farewell cruel world" post is that he says he wants to know "where this factor comes from", and yet he in unable to articulate what he means by this question. He began by saying he would be satisfied if someone could give him the physical law, not involving a transformation, but then when you provided that law he shifted his ground, and began asking where that law "comes from".

Obviously that question is so vague as to be meaningless,

I think you are correct about that, although I do hope that he decides not to give up. He had an unusually high number of people giving him very similar responses, so even if he completely rejects my recent "controversial" example he could still learn a lot just from the commonalities among the responses.

 

[A.T.]

As that's a bit off-topic, just this for the record: That is a myth[2].
The second postulate is a physical law which adheres to Maxwell's wave model and rejects ballistic emission theory; it is defined for a single reference frame. See:
[1] http://www.fourmilab.ch/etexts/einstein/specrel/www/
[2] http://www.uio.no/studier/emner/matnat/fys/FYS-MEK1110/v06/MythsSpecRelativAJP193.pdf

The source independence alone can be stated for a single reference frame, and would qualify as a "physical law" in the sense of the 1 postulate. But that this source independent speed is the same across all inertial frames, is not a "physical law" in that sense.

 

[Samshorn]

The lab analysis was perfectly fine (a is a completely valid parameterization of the worldline in both cases...

No it isn't. You defined "a" as the number of laps that the particle has made around the cyclotron, and T as the period for one lap (in terms of the lab coordinate time t), but when the particle is at rest, the parameter "a" is degenerate (it doesn't change along the particle's worldline), and T is infinite. In essense, you defined a = t/T, so "a" is just a re-scaled version of the coordinate time t, except that it's degenerate when T is infinite (as it in for the muons at rest).

I didn't integrate gamma (if I had then I would have gotten the wrong number since the limits of integration were 0 to 1 instead of 0 to T)...

Huh? Your limits of integration were 0 to a. (By the way, that's the fifth problem with your analysis, using "a" to signify both a parameter and a limit of integration of that parameter, but never mind.) But since "a" doesn't change for the particle at rest, it obviously makes no sense. Also, as has been made crystal clear, you did integrate gamma. You just claim you didn't because instead of using the greek symbol you wrote out the definition of gamma explicitly, and then integrated it (trivially, since it's constant in the cases you considered).

The choice of coordinates is essential since gamma is a feature of a specific set of coordinates (Minkowski)...

No it isn't. The "gamma" expression is just a function of velocity, dS/dt, it doesn't care what space coordinates you are using.

dS/dt was not a part of my analysis.

Yes it was. Your answer for the cyclotron was dtau/dt = sqrt[1-(dS/dt)^2]

 

[Austin0]

):$$c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$$$$\frac{d\tau^2}{dt^2} = 1 - \frac{1}{c^2}\left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right)$$$$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}=\gamma^{-1}$$

Most math literate people work this out pretty quickly - but it should be taught emphatically and early in all SR courses. The metric defines gamma, and in my view comes before gamma logically. It's the metric that welds space and time together, all the other stuff follows.

I think there is an alternative viewpoint;
Gamma is a fundamental aspect of reality.
Aliens traveling alone in an isolated inertial system for a thousand generations might have no concept of other coordinate frames but would still inevitably discover gamma and it would have exactly the same meaning ,describe the same relationship between time,space and velocity, no matter what their coordinate system. Do you think it could be otherwise??

It is not as if there is an arbitrary selection of possible metrics. Any rational (producing accurate predictions) metric must incorporate and correspond to this relationship.
It seems to me that this limits the choices to (+,-,-,-) or (-,+,+,+) for any chosen coordinate system. Is this incorrect??

I don't know GR but aren't there aspects of the EFE , G or others that restrict the possible forms of applicable metrics??

" It's the metric that welds space and time together"
isn't it the gamma function specifically that exactly describes the relationship of space and time??That intrinsically describes and determines the metric. The fundamental spacetime distance measurement.

 

[Dale]

We are going around in circles. I am only responding to the new stuff. Clearly neither of us finds the other's arguments convincing when we repeat them like that.

Your limits of integration were 0 to a.

Oops, that was a mistake that got copy and pasted through.

The "gamma" expression is just a function of velocity, dS/dt, it doesn't care what space coordinates you are using.

Yes it was. Your answer for the cyclotron was dtau/dt = sqrt[1-(dS/dt)^2]

But I never calculated ANYTHING/dt, so I clearly didn't calculate dS/dt. a≠t

 

[Austin0]

The metric did not appear out of thin air and how could it?

In GR, often times metrics do appear "out of thin air". I.e. it is often easier to find a solution to the EFE by starting with the metric and then solving for the distribution of mass and energy required to make that metric.

Well my knowledge of GR is limited but aren't there inherent factors in the EFE that limit and determine the possible applicable metrics?
In any case this is SR

DO you think that gamma could have a different equation giving different relationships?
Could the metric have a different form and still correspond to the gamma relationship of space and time?

The relativistic Doppler equation does not have an explicit gamma function either but would you say that the gamma was not implicit in the math?

You didn't answer the actual question A

The actual question was based on a flawed analogy. So I tried to de-analogize it and answer the underlying question, which I though was a valid question despite the bad analogy.

I asked a question , I did not make any analogy. I am still curious what you thinkk?
Maybe you might wait until I do make an analogy before you declare it flawed.

As far as where it might be lurking, I suspect that harrylin might be on the right track.
the Pythagorean operation returns the value of a line interval in Minkowski geometric space.
... So as the pythagorean operation does perform this transformation, it is in effect a geometric gamma function.

First, I disagree, and second, I never used the Pythagorean theorem nor the Minkowski metric. You are not getting any closer to showing where I used gamma in my calculations.

What do you disagree with?
that the line element is based on the pythagorean theorem?
Or that the theorem doesn't apply using cylindrical coordinates?

More importantly; as you agree that the result contains the gamma factor and I am sure you would agree that this factor was not implicit in the raw coordinate values, the question becomes ,where else could it possibly be lurking??

It lurks in differeniating the line element wrt coordinate time in Minkowski coordinates. I neither used Minkowski coordinates nor did I differentiate wrt coordinate time. You cannot get from what I did to gamma without transforming to Minkowski coordinates and differentiating wrt coordinate time, neither of which are required for calculating the decay of the muons.

$d\tau^2=dt^2-dr^2-r^2d\theta^2-dz^2$ and, as mentioned in post 37, $\tau_P=\int_P d\tau$

dt (untransformed coordinate time) =====> {BLACK BOX} ======> dtau (transformed time -by the gamma factor)
No matter what may have occurred in the black box, Quija board,quantum computer, whatever, that makes no difference.
It is clear that the final value is related to the initial value by the gamma factor?
do you disagree with this statement?

Don't you consider the invariant interval equation a Lorentz transformation??

Well as it seems clear it is a transformation

Nonsense. A transformation requires at least two different coordinate systems and a mapping between them. The metric doesn't even require coordinate systems, let alone a pair of them. So claiming that it a transformation is wrong, and claiming that it is clearly a transformation is absurd.

Perhaps I am not understanding correctly (quite possible) but to me the term itself , invariant interval necessarily implies a multiplicity of frames to have any meaning.
It also didn't seem to apply within a single frame as time intervals are automatically proper time within that frame.
So i understood it to be turning coordinate intervals between events occurring relative to another frame F into a form/value that would agree with all other frames evaluation of that interval relative to F. ------> F's proper time.
Is this incorrect??

It also seems to me that within a frame the metric is still Galilean/Euclidean.
With the assumption of synchronicity of all clocks within the system.
Functionaly equivalent to absolute time.
So the Minkowski metric actually only applies to other frames or is this not so?

 

[zonde]

The source independence alone can be stated for a single reference frame, and would qualify as a "physical law" in the sense of the 1 postulate. But that this source independent speed is the same across all inertial frames, is not a "physical law" in that sense.

Simply reread Einstein's paper http://www.fourmilab.ch/etexts/einstein/specrel/www/:
"... and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body."

There is nothing about speed of light being the same in all inertial frames.

You say:

The 2nd postulate is not just that speed of light is the same in all inertial frames, but also that it is independent of the source velocity in all inertial frames.

Correct statement would be:
"The 2nd postulate is not that speed of light is the same in all inertial frames, but that it is independent of the source velocity."


P.S. Thanks harrylin for pointing out that misconception.

 

[Austin0]

Simply reread Einstein's paper http://www.fourmilab.ch/etexts/einstein/specrel/www/:
"... and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body."

There is nothing about speed of light being the same in all inertial frames.

You say:

Correct statement would be:
"The 2nd postulate is not that speed of light is the same in all inertial frames, but that it is independent of the source velocity."P.S. Thanks harrylin for pointing out that misconception.

Well I have to admit to having that misconception. I could swear i read it in one of the versions of the postulates. But even so it is somewhat implicit in the 1st postulate if you assume that it is inherent in Maxwells equations and fundamental to electrodynamics being the same in all frames , yeah?
or is that a stretch?

 

[Samshorn]

We are going around in circles. I am only responding to the new stuff. Clearly neither of us finds the other's arguments convincing when we repeat them like that.

But you haven't provided any argument at all, let alone repeated one. You simply defined T as the time it takes a stationary particle to make one lap around a cyclotron, and I pointed out that a stationary particle will never complete a lap around a cyclotron, so T is infinite, and this invalidates your derivation. So far, your only answer has been... well... nothing.

... the limits of integration were 0 to 1 instead of 0 to T...

samshorn: Huh? Your limits of integration were 0 to a.

Oops, that was a mistake that got copy and pasted through.

Now that this has been pointed out, can you clarify what integration limits you intended? According to what you said above, you intended to integrate over "a" from 0 to 1, but that doesn't make any sense, because the parameter "a" appears in your results. But 0 to T wouldn't make any sense either. You must have intended to integrate over a dummy angular variable from 0 to "a", right? But if so, what exactly does that dummy variable represent for the stationary particle? It can't be angular position, because the angular position of a stationary particle doesn't change.

 

[harrylin]

[..] That this source independent speed is the same across all inertial frames, is not a "physical law" in that sense.

Indeed it's not. While the constancy of light speed in an inertial frame is a physical law, that invariance of physical laws between inertial frames is imposed by the PoR which could be called a kind of "meta-law" - a law about physical laws.

 

[harrylin]

[..]
dt (untransformed coordinate time) =====> {BLACK BOX} ======> dtau (transformed time -by the gamma factor)
No matter what may have occurred in the black box, Quija board,quantum computer, whatever, that makes no difference.
It is clear that the final value is related to the initial value by the gamma factor? [..]
Perhaps I am not understanding correctly (quite possible) but to me the term itself , invariant interval necessarily implies a multiplicity of frames to have any meaning. [..]
So the Minkowski metric actually only applies to other frames or is this not so?

Well seen! I agree with most of your and Samshorn's arguments. One can use the Lorentz transformation or its equivalent, the space-time interval, to discover the physical law (=in a single frame) about the slowdown of moving radioactive clocks by a factor of gamma. And from there on one can apply the thus found physical law without help of system transformations, as Einstein also did in his 1905 paper ("It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line. [..] assume that the result [..] is also valid for a continuously curved line").

 

[Dale]

But you haven't provided any argument at all, let alone repeated one. You simply defined T as the time it takes a stationary particle to make one lap around a cyclotron, and I pointed out that a stationary particle will never complete a lap around a cyclotron, so T is infinite, and this invalidates your derivation. So far, your only answer has been... well... nothing.

T is the time that it takes one of the cyclotron particles to go around. It is the same value for both worldlines. I didn't re-calculate T for the lab particles. I.e. it is the cyclotron period, for both sets of particles. Similarly R is also the radius of the cyclotron for both sets of particles even though the lab particles are not going around the cyclotron.

Regardless of the text, it should have been abundantly clear just from inspection of the equations themselves that both $(t,r,\theta,z)=(aT,R,a2\pi,0)$ and $(t,r,\theta,z)=(aT,R,0,0)$ are valid parametric equations, and that the first correctly represents the worldline of a cyclotron particle parameterized by a and the second correctly represents the worldline of stationary particles parameterized by a, and that the two worldlines intersect at a=0 and a=1.

Btw, I just realized that a cannot be a re-scaled t. The units are wrong. t has units of time and a is unitless. Similarly, dS/da has units of distance while dS/dt has units of speed. And dτ/da has units of time while dτ/dt is unitless.

Now that this has been pointed out, can you clarify what integration limits you intended? According to what you said above, you intended to integrate over "a" from 0 to 1, but that doesn't make any sense, because the parameter "a" appears in your results. But 0 to T wouldn't make any sense either. You must have intended to integrate over a dummy angular variable from 0 to "a", right? But if so, what exactly does that dummy variable represent for the stationary particle? It can't be angular position, because the angular position of a stationary particle doesn't change.

I had originally intended to integrate from 0 to 1 (I was thinking of it as a twin's scenario with the twins separating at a=0 and reuniting at a=1). In any case, it is a very minor mistake, and perhaps using a as the limit of integration is better anyway.

If you want to pick on such minor notational abuses (presumably because you don't have any substantative criticisms) then you should probably also point out that I shouldn't use the same variables P and a for two different worldlines, but should have used different variables for each. I am sure there are other similarly minor mistakes.

 

[Dale]

Well my knowledge of GR is limited but aren't there inherent factors in the EFE that limit and determine the possible applicable metrics?

Not really, I mean, you can pretty much find a stress energy tensor that fits any metric as far as I know. That is why you can have things like wormholes which are valid solutions to the EFE but which are not considered "physical".

Generally people add "energy conditions" to distinguish between physical and non-physical solutions to the EFE, but the energy conditions are put in ad hoc and are not inherent to the EFE.

DO you think that gamma could have a different equation giving different relationships?
Could the metric have a different form and still correspond to the gamma relationship of space and time?

I don't know what you mean by "the gamma relationship of space and time". Gamma is a particular expression: $\gamma=(1-v^2/c^2)^{-1/2}$. It is easy to get this expression from the Minkowski metric. As far as I know, it is not possible to get it from other metrics without doing a transform to a (local) Minkowski coordinate system. Although Samshorn's point that v=dS/dt does expand the class of metrics where you can get it.

$d\tau^2=dt^2-dr^2-r^2d\theta^2-dz^2$ and, as mentioned in post 37, $\tau_P=\int_P d\tau$

dt (untransformed coordinate time) =====> {BLACK BOX} ======> dtau (transformed time -by the gamma factor)
No matter what may have occurred in the black box, Quija board,quantum computer, whatever, that makes no difference.
It is clear that the final value is related to the initial value by the gamma factor?
do you disagree with this statement?

There is no black box which takes dt as input and gives dτ as output in general. For certain specific problems (constant speed particles) in specific metrics (Minkowski) you can simplify the inputs to the black box as you describe. But in general the inputs to the black box are all of the dx_i, not just dt, and in fact in some metrics there is no dt to begin with.

Of course, I do agree that any physical experiment where the outcome is a function of gamma will have the same outcome regardless of the metric used to analyze the experiment. The point I was making is that the correct outcome can be obtained by simple application of the law as stated, without at any point explicitly bringing gamma into the analysis nor doing any transforms.

Perhaps I am not understanding correctly (quite possible) but to me the term itself , invariant interval necessarily implies a multiplicity of frames to have any meaning.

True. The invariant interval is also known as the spacetime interval, the line element, and the metric. None of which imply another frame. I probably should have used the term "line element".

It also didn't seem to apply within a single frame as time intervals are automatically proper time within that frame.

This is not true in general. For example, consider a rotating reference frame. For large values of r the t coordinate is spacelike.

So i understood it to be turning coordinate intervals between events occurring relative to another frame F into a form/value that would agree with all other frames evaluation of that interval relative to F. ------> F's proper time.
Is this incorrect??

I am not certain that I understand what you are saying, but I don't think that is incorrect, just a bit "cumbersome".

Relativity, particularly GR, is a geometric theory. Just think about the spacetime interval as a strange "distance" in spacetime. Distance is a geometric feature, it exists independently of any coordinate system that you might draw on top of the geometric features.

It also seems to me that within a frame the metric is still Galilean/Euclidean.
With the assumption of synchronicity of all clocks within the system.
Functionaly equivalent to absolute time.

Not in general, particularly for non-inertial reference frames.

So the Minkowski metric actually only applies to other frames or is this not so?

This is not so. It applies to every inertial frame, not just "other" inertial frames.

 

[Mentz114]

So the Minkowski metric actually only applies to other frames or is this not so?

If we boost the Minkowski metric by β in the x direction the interval ds² = dt² - dx² - dy² - dz² becomes

ds'² = (γ dt + βγ dx)² - (γ dx + βγ dt)² - dy² - dx²
= γ²(1-β²) dt² - γ²(1-β²) dx² -dy² - dx²
= dt² - dx² - dy² - dz²

It is invariant under LT.

 

[A.T.]

Well I have to admit to having that misconception. I could swear i read it in one of the versions of the postulates.

According to:
http://www.uio.no/studier/emner/matnat/fys/FYS-MEK1110/v06/MythsSpecRelativAJP193.pdf
this "constant across frames" version is the more popular now:

Q. Does the phrase, “the constancy of the speed of light,”
have the same meaning today that it had when Einstein used
it in 1905?
A. No.
Today, the primary meaning of the phrase is that, given a
specific burst of light, the burst’s speed is measured to have
the same numerical value in all inertial frames. That is, the
speed is constant with respect to changes in the reference
frame in which it is observed.
A secondary meaning also exists: in any given frame,
bursts of light from sources with different velocities all have
the same speed. That is, the speed of light is constant with
respect to changes in the source’s velocity.

But I agree with the author that the original "independent of source" version is better, because it doesn't invoke the wrong idea, that the 2nd postulate is somehow redundant. The constancy across frame then follows from the combination of 1 & 2 postulate.

 

[stevendaryl]

Perhaps I am not understanding correctly (quite possible) but to me the term itself , invariant interval necessarily implies a multiplicity of frames to have any meaning.

Not really. Let me give an analogy. Take a piece of paper, and make two dots on it, points A and B. Draw a curve connecting the two points. That curve has a length, and the length has a meaning that is independent of which coordinate system you use to compute the length. We can represent that length in the following way:

Divide the curve into tiny little segments by selecting a bunch of points along the curve P₀, P₁, ... P_N. Then define the vector V_i to be the line connecting P_i to P_i+1. Then the length of the curve is (in the limit as N → ∞) given by L = sum over all i of |V_i|, where |V_i| is the length of vector V_i. This definition doesn't refer to coordinates at all. However, to compute the length of vector V_i, we can pick a coordinate system (x,y) to describe V_i, and compute the length as
√(δx² + δy²).

The spacetime interval is exactly the same sort of thing. You have two events (points in spacetime) A and B. You have a path connecting the two events. You divide the path into segments by picking events along the curve e_i. Then define the spacetime vector V_i to be the vector connecting e_i to e_i+1. Each vector V_i has a "length", |V_i|, and to compute the "length" of the whole path, you just add up the lengths of the V_i. This definition doesn't refer to coordinates at all. However, to compute the length of vector V_i, we can pick a coordinate system (x,t) to describe V_i, and compute the length as
√(δt² - δx²/c²).

 

[Samshorn]

T is the time that it takes one of the [moving] cyclotron particles to go around. It is the same value for both worldlines. I didn't re-calculate T for the [stationary] lab particles.

EXACTLY. And do you understand that this invalidates all your claims? Remember, I've been telling you from the start that you actually integrated 1/gamma by the coordinate time t, because your "a" parameter for the stationary particle is NOT really the stationary particle's angular position, it is simply a re-scaled coordinate time, a = t/T for an arbitrary constant T that has nothing to do with the stationary particle's worldline.

Now that you realize this, and hence that 'a' does not represent anything other than t/T where T is an arbitrary constant (as far as the stationary particle is concerned), go back and re-do your analysis for the stationary particle, taking everything you've learned (so far) into account. Get rid of the obfuscational 'a' and T, and simply write it honestly in terms of t. You will find that your integral of (dtau/da)da written honestly is simply (dtau/dt)dt, and of course dtau/dt = 1/gamma, so (surprise!) you are just integrating the constant 1/gamma, which is what I've been telling you all along.

I just realized that 'a' cannot be a re-scaled t. The units are wrong. t has units of time and a is unitless.

Utter nonsense. If we define beta = v/c, it is unitless, whereas v has units of velocity, but obviously beta is just a re-scaled v. In fact, by choosing units so c = 1 we often write beta as just v, recognizing that it is a dimensionless version of velocity. You simply cannot deny that your 'a' parameter (for the stationary particle) is a = t/T and cannot even be interpreted as angular position because the particle isn't moving. It can ONLY be interpreted as an arbitrarily re-scaled coordinate time. Once you recognize this, you see that you simply integrated 1/gamma, albeit in obfuscational notation.

If you want to pick on such minor notational abuses (presumably because you don't have any substantative criticisms)...

Now, that is really unfair, isn't it? Your whole substantive point - your whole stated reason for doing the analysis - was to show that the analysis did NOT consist of integrating 1/gamma, and my criticism (which you've just tacitly conceded) is that you DID just integrate 1/gamma. Surely this qualifies as a substantive criticism. The reason I had to point out all the notational abuses is that you used those abuses to try to disguise the 1/gamma integration by using a re-scaled and re-named coordinate time, wrongly claiming that it represents the angular position of the particle, whereas now you have (finally!) admitted that it represents no such thing, and is really nothing but the coordinate time divided by the arbitrary constant T, and hence you were indeed simply integrating 1/gamma.