Transformation Vs. Physical Law

[Dale]

But if you think, that comparing the decay of Muons in lab and in cyclotrons does not involve the gamma factor. Please post some calculations for the same.

OK, so I will use units where c=1, and the decay law in post 13: n=n_0 e^{-\lambda \tau}. Assuming that the initial number and the decay constant are known then all that remains is to calculate tau for two cases, one being muons in a cyclotron and the other being muons at rest in the lab. We will neglect gravity and, since cyclotrons are circular it will be convenient to use cylindrical coordinates. The flat spacetime metric in cylindrical coordinates is: 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

For the first case, the muons in the cyclotron, we can write their worldline as P=(t,r,\theta,z)=(a T, R, a 2 \pi, 0) where a is the number of "laps" around the cyclotron, R is the radius of the cyclotron, and T is the period for one lap. So we have
\tau_P=\int_P d\tau=\int_0^a \frac{d\tau}{da}da=\int_0^a \sqrt{\frac{dt^2}{da^2}-\frac{dr^2}{da^2}-r^2\frac{d\theta^2}{da^2}-\frac{dz^2}{da^2}} \; da=\int_0^a \sqrt{T^2-0-R^2 4 \pi^2-0} \; da=a\sqrt{T^2-4\pi^2 R^2}

For the second case, the muons at rest in the lab, we can write their worldline as P=(t,r,\theta,z)=(a T, R, 0, 0). So we have
\tau_P=\int_P d\tau=\int_0^a \frac{d\tau}{da}da=\int_0^a \sqrt{\frac{dt^2}{da^2}-\frac{dr^2}{da^2}-r^2\frac{d\theta^2}{da^2}-\frac{dz^2}{da^2}} \; da=\int_0^a \sqrt{T^2-0-R^2 0-0} \; da=aT

So, I have calculated the number of decayed particles for particles at rest in the lab and in a cyclotron without using \gamma = (1-v^2)^{-1/2}.

 

[harrylin]

[..] So, I have calculated the number of decayed particles for particles at rest in the lab and in a cyclotron without using \gamma = (1-v^2)^{-1/2}.

Very neat (thanks!) - but if I'm not terribly mistaken it's only half true. With the "flat spacetime metric" you obviously mean the space-time interval which is based on the Lorentz group* - and thus on the gamma factor. So implicitly gamma is hiding in your calculation.

However, (and this is for universal): the fact that the gamma factor is used for transformation equations doesn't mean that it's not an implicit or explicit part of natural law as well - as many of us have tried to explain by now.

*http://en.wikisource.org/wiki/On_the_Dynamics_of_the_Electron_%28July%29#.C2.A7_4._.E2.80.94_The_Lorentz_group

 

[Dead Boss]

I would say that the Minkowski metric is more fundamental than the gamma factor.

 

[Dale]

So implicitly it's hiding in your calculation.

Where, exactly?

Certainly you can derive gamma from the metric, but that doesn't mean that every calculation involving the metric involves gamma. (and also not every calculation involving gamma is a transform, but that is a separate discussion)

 

[Dale]

I would say that the Minkowski metric is more fundamental than the gamma factor.

Agreed. A metric can be defined on a Riemannian manifold independently of any coordinate charts and any transforms between different coordinate charts. Gamma is a factor in a specific type of transform between a specific class of coordinate systems on a flat spacetime, which is several steps less general than the metric on the flat spacetime.

 

[harrylin]

Where, exactly?

Exactly in the section to which I referred, about the rotation angle (it's simply Pythagoras):

it is easy to see that this transformation is equivalent to a coordinate change, the axes are rotating a very small angle around the z-axis. [..] Any transformation of this group can always be decomposed into [..] a linear transformation which does not change the quadratic form x²+y²+z²-t²

Certainly you can derive gamma from the metric, but that doesn't mean that every calculation involving the metric involves gamma. (and also not every calculation involving gamma is a transform, but that is a separate discussion)

I think that the fact that not every calculation involving gamma is a transform, is at the heart of the discussion; that it's the sticking point for universal. And he seems to have not responded to a dozen of posts in which this has been stressed.

 

[Dale]

Exactly in the section to which I referred, about the rotation angle (it's simply Pythagoras):

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.

I think that the fact that not every calculation involving gamma is a transform, is at the heart of the discussion; that it's the sticking point for universal. And he seems to have not responded to a dozen of posts in which this has been stressed.

Yes, he does seem to have a strong tendency to simply ignore posts which refute his argument. Otherwise this thread would be much shorter and his previous thread would not have been locked.

 

[universal_101]

I think that the fact that not every calculation involving gamma is a transform, is at the heart of the discussion; that it's the sticking point for universal. And he seems to have not responded to a dozen of posts in which this has been stressed.

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. And as all of you are suggesting, I should stop questioning since everything has been answered.

And it means this thread can be safely locked.

 

[Dale]

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

Do you have a specific non-transform equation with gamma that you are interested in? I think, for a given equation, we could explain the source, but I am not sure that the answer is the same for all such equations.

 

[Mentz114]

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. And as all of you are suggesting, I should stop questioning since everything has been answered.

And it means this thread can be safely locked.

Maybe what you are calling a gamma factor is a property of the Minkowski spacetime, the background ( or theatre) in which the physical laws act.

 

[Austin0]

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.


Yes, he does seem to have a strong tendency to simply ignore posts which refute his argument. Otherwise this thread would be much shorter and his previous thread would not have been locked.

Isn't it true that the metric itself is the direct result of the gamma function and its implications regarding the relationship between time,space and relative states of motion.

it seems to me that the coordinate system and derivative body of transformation maths is simply the fundamental gamma function applied to and implemented through the Galilean metric.

This is explicitly manifest in the reverse sign of the time term , which of itself necessarilly implies and expresses time dilation.
The ratio of the temporal and spatial componenets describe a velocity and the resulting proper time value is the gamma corrected, dilated time value related to this velocity.

To me this appears to be a case of math magic, integrating the function so that the gamma calculation is not necessary as a separate operation. As is the case with the Doppler formula.

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?
That to apply the formula was not indirectly applying that function?

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

your thoughts?

 

[Austin0]

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. And as all of you are suggesting, I should stop questioning since everything has been answered.

And it means this thread can be safely locked.

The gamma function comes directly from the intrinsic properties of the physical world.
It is simply a description of the fundamental relationship of time, space and motion. If it was not discovered through Maxwell it would have been through particle accelerations or other empirical measurements. So you have it backwards. The function does not come from the transformation; The transformation comes from the function. And that was always there

 

[Dale]

wow Austin0, that is a lot of questions. My answers will be necessarily brief, but if you want to go deeper into one or two, I will be glad to.

Isn't it true that the metric itself is the direct result of the gamma function and its implications regarding the relationship between time,space and relative states of motion

No. The metric is more fundamental than any coordinate system or transform.

This is explicitly manifest in the reverse sign of the time term , which of itself necessarilly implies and expresses time dilation.
The ratio of the temporal and spatial componenets describe a velocity and the resulting proper time value is the gamma corrected, dilated time value related to this velocity.

This is correct when the metric is expressed in terms of an inertial coordinate system in flat spacetime, but not in general.

To me this appears to be a case of math magic, integrating the function so that the gamma calculation is not necessary as a separate operation. As is the case with the Doppler formula.

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?

I have asked several times for people to show me where exactly the gamma was implicit in my formulas. I don't know where it is supposed to be lurking.

That to apply the formula was not indirectly applying that function?

Huh?

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

No.

 

[Samshorn]

I have asked several times for people to show me where exactly the gamma was implicit in my formulas. I don't know where it is supposed to be lurking.

This was explained in post #114. Your formula is \tau_P=\int_P d\tau where d\tau^2=dt^2-dr^2-r^2d\theta^2-dz^2, but the latter can be written as d\tau^2=dt^2 / \gamma^2, so you're just integrating a function of gamma. Of course, as explained in #114, written in this form we are essentially computing the ratio of proper time to coordinate time, which is ultimately what we always do. Gamma is a function of the velocity in terms of our chosen coordinate system with the time coordinate t. If we chose a different coordinate system, t would be different and so would gamma. But as explained in #114, the number of muons at any event in invariant. But the elapsed times in two different frames are related by gamma, as confirmed by the ratio of your two answers. Normally we combine your two calculations into just one for the ratio, since that's what we ultimately care about, and hence gamma appears.

This is correct when the metric is expressed in terms of an inertial coordinate system... but not in general.

If both reference systems are accelerating then the comparison of elapsed times would be expressible in terms of the combined effect of two "gammas". Of course, the ratio of differental elapsed times can always be expressed in terms of a single "gamma".

 

[bobc2]

The metric is more fundamental than any coordinate system or transform...

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

 

[Dale]

This was explained in post #114. Your formula is \tau_P=\int_P d\tau where d\tau^2=dt^2-dr^2-r^2d\theta^2-dz^2, but this can be written as

d\tau^2=\frac{dt^2}{\gamma^2}

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

If you do a transform and differentiate wrt coordinate time as you explained in 114, then you can indeed derive gamma. I have mentioned that before. However, I neither transformed to an appropriate coordinate system nor did I differentiate wrt coordinate time. I didn't use gamma either implicitly nor explicitly, although you certainly can branch off from what I did and derive gamma.

so you're just integrating a function of gamma. Of course, as explained in #114, written in this form we are essentially computing the ratio of proper time to coordinate time, which is ultimately what we always do.

No, in my case it was the ratio of proper time to the parameter a. If I had parameterized the worldline by t then that would be the case. Often, proper time itself is used as the parameter, so you just integrate 1. There is never any need to use t as the parameter.

If both reference systems are accelerating then the comparison of elapsed times would be expressible in terms of the combined effect of two "gammas". Of course, the ratio of differental elapsed times can always be expressed in terms of a single "gamma".

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

 

[Austin0]

Isn't it true that the metric itself is the direct result of the gamma function and its implications regarding the relationship between time,space and relative states of motion

No. The metric is more fundamental than any coordinate system or transform.

you may be right about the abstract question of general fundamentality
but this is about a specific metric.
I appears to me that this metric could not exist without the basic understanding of the relationship of space and time described by Lorentz in the gamma function. The metric did not appear out of thin air and how could it?

This is explicitly manifest in the reverse sign of the time term , which of itself necessarily implies and expresses time dilation.
The ratio of the temporal and spatial components describe a velocity and the resulting proper time value is the gamma corrected, dilated time value related to this velocity.

This is correct when the metric is expressed in terms of an inertial coordinate system in flat spacetime, but not in general.

Good , so we are agreed that the result does contain the gamma factor.

To me this appears to be a case of math magic, integrating the function so that the gamma calculation is not necessary as a separate operation. As is the case with the Doppler formula.

A 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?

I have asked several times for people to show me where exactly the gamma was implicit in my formulas. I don't know where it is supposed to be lurking.

You didn't answer the actual question A

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. The geometry of that space seems to me to intrinsically require the gamma factor to transform the geometry regarding the moving system into meaningful quantities.

So as the pythagorean operation does perform this transformation, it is in effect a geometric gamma function.

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??

Pure logic leads to the inevitable conclusion that it must necessarily be implicit,(hiding) in the operation leading to that result. Unless you think it is just coincidence ;-)

Would you say That to apply the Doppler formula was not (indirectly) applying that function?

Huh?

* Sorry if my phrasing was a little ambiguous. Does this track better??

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

No.

Well as it seems clear it is a transformation does this mean you don't think it is part of the general derivative functions stemming from the original math?

 

[Nugatory]

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

That's not so much something that can be established/proven as it is an argument from aesthetic principles; if the theory is internally consistent I can start at either end and get to the other, so neither end is demonstrably more fundamental.

But I'd expect that this aesthetic judgement is shared by the overwhelming majority of people who have figured out the underlying math and physics. The metric describes properties of space-time that exist independent of any coordinate system; the coordinate systems, reference frames, and transformations between them were invented by us to make it easier to map these properties to our own experience.

 

[ImaLooser]

Thanks for the view,

I agree that Lorentz transformation is more than just a transformation in modern physics. It is exactly what I'm questioning. It seems as if the transformation is multipurpose, it can be a physical law at times and also can be a transformation at other.

Do you see this contradiction of basic physics concept.

What is the difference between a physical law and a transform? Not much. Hard to say, really. The boundary is vague. I wouldn't worry about it. If you want to call the Lorentz transform a physical law, that is fine with me.

 

[Nugatory]

I appears to me that this metric could not exist without the basic understanding of the relationship of space and time described by Lorentz in the gamma function. The metric did not appear out of thin air and how could it?

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.

 

[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.