Transformation Vs. Physical Law

[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.

 

[Dale]

And do you understand that this invalidates all your claims?

Nonsense, it doesn't invalidate anythying.

it is simply a re-scaled coordinate time, a = t/T

a=t/T is unitless, it does not have units of time. No matter how many times you assert that it is time, the assertion is false.

Now that you admit that this, and hence that 'a' does not represent anything other than t/T where T is an arbitrary constant

I never said otherwise. In fact, for the lab particles I very clearly said (t,r,\theta,z)=(aT,R,0,0) so obviously I intended from the beginning for t=aT. For the cyclotron particles (t,r,\theta,z)=(aT,R,a2\pi,0) so t=aT and θ=a2π. In all cases, a is nothing more nor less than a unitless variable which parameterizes the worldline.

go back and re-do your analysis ... Get rid of the obfuscational 'a' and T

And that is my point. In order to make your claim you have to go back and re-do the analysis differently from how I did it. I never said that you couldn't derive gamma (in fact I said the opposite many times), I only said and demonstrated that it is not necessary to do so. Certainly, if I went back and did the analysis differently than I did, I could come up with gamma. That doesn't alter the fact that doing the analysis the way I did never used gamma, even in disguise. To make that claim you are having to go back and do a different (but equivalent) analysis than what I did.

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.

Beta is unitless, v has units of speed, therefore beta is not a re-scaled v. If beta were a rescaled v then it would have units of speed and you could not subtract it from 1.

In fact, by choosing units so c = 1 we often write beta as just v, recognizing that it is a dimensionless version of velocity.

Sure, with the understanding that you can always add some power of c back into make the units work out. There is no power of c that you can add to a in order to get it in units of time.

Now, that is really unfair, isn't it?

And so is continuing to harp on a minor notational mistake when:
1) I admitted immediately that it was a mistake
2) it makes no difference at all

 

[Samshorn]

In order to make your claim you have to go back and re-do the analysis differently from how I did it.

Well, you obviously need to fix all the problems and remove all the misconceptions that have been identified in your analysis (logically inconsistent variable definitions, invalid integration limits...). Without fixing these errors, your 'analysis' is just gibberish. Once you have fixed them, you will see that the analysis simply consists of integrating 1/gamma by coordinate time dt. Remember, your whole point was that the analysis doesn't integrate over the coordinate time dt, but we've seen that you "accomplished" this merely by changing the symbol for coordinate time from t to a and then integrating over a. Obviously it's absurd to claim that you are not integrating over coordinate time, simply because you changed the symbol for coordinate time from t to a. And the choice of units for coordinate time is obviously irrelevant (dividing by the arbitrary scale factor T).

a=t/T is unitless, it does not have units of time.

Again, it's irrelevant what units we use for coordinate time, it is still coordinate time. Dividing by an arbitrary scale factor obviously doesn't change that.

 

[Dale]

Obviously it's absurd to claim that you are not integrating over coordinate time, simply because you changed the symbol for coordinate time from t to a.

If a were merely a change of symbol for t then we would have t=a, but we don't. We have t=Ta≠a.

And the choice of units for coordinate time is obviously irrelevant (dividing by the arbitrary scale factor T).

So you think it would be OK to use units of Newtons for coordinate time, or perhaps pascals? If so then why, if not then why not?

 

[Samshorn]

So you think it would be OK to use units of Newtons for coordinate time, or perhaps pascals? If so then why, if not then why not?

The quantity t/T where T is a constant (which you defined as the time it takes for an arbitrarily selected reference muon to complete 1 lap in a cyclotron) represents coordinate time in units of "cyclotron reference lap times". That's a weird choice of units, but it's just as valid as the unit based on the standard second (which is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom). Whenever we assign a numerical value to a time interval it is always a ratio to some other reference time interval, whether it be a standard second or a "cyclotron reference lap time" or any other defined reference interval.

I seriously urge you to actually perform the calculation, rather than just talking about it. Try eliminating all the errors and misconcentions that we've identified in your earlier attempt. Explain what parameter you are integrating for the stationary particle, and why. It obviously isn't angular position, as you claimed in your original attempt, so what is it? And what is your limit of integration? Until you've actually done the calculation, I don't think just talking about it more (or posing obtuse questions about the meaning of physical units) is going to help.

 

[Dale]

The quantity t/T where T is a constant (which you defined as the time it takes for an arbitrarily selected reference muon to complete 1 lap in a cyclotron) represents coordinate time in units of "cyclotron reference lap times".

No, it isn't. The quantity aT is equal to t in units of "laps", not a.

You can measure t in units of seconds, or laps, or fortnights, whatever units you use it has dimensions of time. You can measure T in units of seconds, or laps, or fortnights, whatever units you use it has dimensions of time. Therefore the ratio t/T=a is dimensionless. A dimensionless quantity cannot be equal to a quantity with dimensions of time any more than a quantity in units of Newtons could.

I am sorry that you are finding the units so difficult here. My first day of my first physics class the instructor spent the entire class time hammering into us the importance of always checking units at each step. It was surprisingly memorable, and has proven to be the single most useful piece of physics advice I ever received. It saved my grade multiple times in multiple classes. I am sorry that you didn't receive a similar educational experience.

 

[Samshorn]

The quantity aT is equal to t in units of "laps", not a.

No, for the stationary particle the quantity 'a' does not represent any angular measure of laps or anything like that, because the particle is stationary. Its angular position is not changing. You yourself have admitted that 'a' is nothing but a parameterization of the worldline of the particle, and its only definition is a = t/T where T is just an arbitrary unit of time that you carried over (unwittingly) from a different case. So 'a' is nothing but coordinate time expressed in units of "reference particle lap times". Needless to say, the units of "a" don't matter, because you're multiplying dtau by da/da. So you just need to be sure the limits of integration are specified in consistent units with the variable of integration.

Look, the relevant integrand is (dtau/dt)dt = dt/gamma, but you choose to write this in the more convoluted form (dtau/d[t/T])d[t/T], and you claim with a straight face that, because you've written it this way, replacing t with t/T for some arbitrary constant T, it no longer involves gamma or the coordinate time t, not even implicitly, despite the fact that the T's obviously cancel out, leaving nothing but dt/gamma.

I honestly don't see how you can seriously claim that, simply by multiplying dt/gamma by T/T for some arbitrary constant T, the resulting expression no longer involves dt/gamma, not even implicitly!

 

[Dale]

And I honestly don't see how you can seriously claim that a dimensionless quantity has dimensions of time.

 

[GAsahi]

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}.

This is not quite right, since the muons move in a cyclotron, r=R so dr=dz=0 meaning that d\tau^2=dt^2-r^2d\theta^2=dt^2(1-R^2 (\frac{d \theta}{dt})^2) meaning that d\tau^2=dt^2 (1-v^2)=(\frac{dt}{\gamma})^2. There is really no reason to introduce the number of laps, a as a variable of integration, the integration variable is the coordinate time t. This is the standard for problems of this type.

 

[Samshorn]

And I honestly don't see how you can seriously claim that a dimensionless quantity has dimensions of time.

It's quite common in relativity to normalize all variables so that everything (distances, times, masses,...) all have consistent units, such as all having units of meters (so-called geometric units). In these units, the mass of the Sun is 1.475 kilometers. That's right... the mass of the Earth has units of distance. Gasp! And we can just as well divide all these quantities (distances, times, masses,...) by some arbitrary reference distance (such as the diameter of your navel), to make them all dimensionless. These are perfectly fine units. Of course, what we've really done is converted everything to units of "Dalespam's navel diameters".

Again, every numerical value of a physical quantity is a ratio of the variable to some reference value along with arbitrary scale factors - this is the meaning of units. What your high school teacher tried (and evidently failed) to teach you is that we must have consistency of units, not that we are obligated to use any particular units (e.g., metric versus English, or seconds versus Dalespam navels). Of course, in the case under discussion, where the coordinate time appears only as dt/dt, it isn't even necessary for the units of t to be the same as the units of tau. All we need to ensure is that the units are consistent with the limits of integration. But in more general circumstances we are still always free to define whatever units we like, provided only that we are consistent.

So, now that we've cleared up your misunderstanding of elementary units, hopefully you can see that your 'a' is nothing but t in weird units, and your analysis simply consisted of integrating dt/gamma.

 

[Dale]

There is really no reason to introduce the number of laps, a as a variable of integration

There is really no reason not to either. Reparameterizations are common and well accepted.

 

[Dale]

It's quite common in relativity to normalize all variables so that everything (distances, times, masses,...) all have consistent units, such as all having units of meters (so-called geometric units). In these units, the mass of the Sun is 1.475 kilometers. That's right... the mass of the Earth has units of distance.

In geometrized units t has units of length. It is still not dimensionless.

 

[Samshorn]

In geometrized units t has units of length. It is still not dimensionless.

Your misunderstanding of what "dimensionless" means was explained in the very next sentence after the one you quoted:

And we can just as well divide all these quantities (distances, times, masses,...) by some arbitrary reference distance (such as the diameter of your navel), to make them all dimensionless. These are perfectly fine units. Of course, what we've really done is converted everything to units of "Dalespam's navel diameters".

Again, every numerical value of a physical quantity is a ratio of the variable to some reference value along with arbitrary scale factors - this is the meaning of units. What your high school teacher tried (and evidently failed) to teach you is that we must have consistency of units, not that we are obligated to use any particular units (e.g., metric versus English, or seconds versus Dalespam navels). Of course, in the case under discussion, where the coordinate time appears only as dt/dt, it isn't even necessary for the units of t to be the same as the units of tau. All we need to ensure is that the units are consistent with the limits of integration. But in more general circumstances we are still always free to define whatever units we like, provided only that we are consistent.

So, now that we've cleared up your misunderstanding of elementary units, hopefully you can see that your 'a' is nothing but t in weird units, and your analysis simply consisted of integrating dt/gamma.

 

[Dale]

Your misunderstanding of what "dimensionless" means was explained in the very next sentence after the one you quoted:

And we can just as well divide all these quantities (distances, times, masses,...) by some arbitrary reference distance (such as the diameter of your navel), to make them all dimensionless. These are perfectly fine units. Of course, what we've really done is converted everything to units of "Dalespam's navel diameters".

You seem to be missing some basic algebra. If you start with geometrized units t is in e.g. meters and a is dimensionless. If you then "divide all these quantities" by navels then t' will be dimensionless but a' will have units of navels^-1. There is no factor that you can divide out to get t and a into the same units. They are dimensionally inconsistent no matter how much you wish otherwise.

 

[Samshorn]

If you start with geometrized units t is in e.g. meters and a is dimensionless. If you then "divide all these quantities" by navels then t' will be dimensionless but a' will have units of navels^-1. There is no factor that you can divide out to get t and a into the same units.

Of course they have different units. That's what I've been telling you. The variables t and 'a' both represent coordinate time, but in different units. To argue that they can't both represent coordinate time because they have different units is absurd. Hours, seconds, reference lap times... the units don't change what the variable represents, not even if you choose to give the variable different names in different units.

You defined 'a' as coordinate time in units of "reference lap times", which is a pure number and hence dimensionless, but it still represents coordinate time. Just as 'beta' (=v/c) and v both represent velocity, your variable 'a' (=t/T) and t both represent coordinate time. Yes, they have different units. But obviously this does not imply that beta (which is numerically equal to v if we choose units such that c=1) does not represent velocity. And it does not imply that 'a' (which is numerically equal to t if we choose units such that T=1) does not represent coordinate time. It's purely a matter of arbitrary units (and even more arbitrary in your case, since your scale factor T is itself totally arbitrary).

All this merely confirms that you just unwittingly integrated dt/gamma, and confused yourself into thinking you had done something different by using a different symbol for coordinate time in dimensionless units.

 

[Dale]

Of course they have different units. That's what I've been telling you. The variables t and 'a' both represent coordinate time, but in different units. To argue that they can't both represent coordinate time because they have different units is absurd. Hours, seconds, reference lap times... the units don't change what the variable represents, not even if you choose to give the variable different names in different units.

Perhaps you don't understand the difference between units and dimensions. t and a not only have different units, they have different dimensions. That means that there is no conversion factor between them. Here is a good place to start:

http://en.wikipedia.org/wiki/Dimension_of_a_physical_quantity

You defined 'a' as coordinate time in units of "reference lap times", which is a pure number and hence dimensionless, but it still represents coordinate time.

Amazing. You really seem to honestly believe that a dimensionless number nonetheless has dimensions of time.

 

[Samshorn]

Perhaps you don't understand the difference between units and dimensions. t and a not only have different units, they have different dimensions.

No, you're still confused. Take a look at
http://www.owlnet.rice.edu/~labgroup/pdf/Dimensions_units.pdf

In summary, it says:
"It's easy to confuse the physical dimensions of a quantity with the units used to measure the dimension. We usually consider quantities like mass, length, and time as fundamental dimensions, and then we express the dimensions of other quantities like speed, which is length/time, in terms of the basic set. Every quantity which is not explicitly dimensionless, like a pure number, has a characteristic dimension which is not affected by the way we measure it. Units give the magnitude of some dimension relative to an arbitrary standard. For example, when we say that a person is six feet tall, we mean that person is six times as long as an object whose length is defined to be one foot."

Do you understand this? When we say a person is six feet tall, we are giving the ratio of two lengths (the person and the reference object), so the ratio is dimensionless, but we still say the person's height has the dimension of length, and it's magnitude has the dimensionless value 6 in units of feet. The magnitude of the length in any specified units is dimensionless.

This confuses people because they don't understand the degree to which our terminology is conventional, and because they don't clearly distinguish between variables and reference constants. Think about spatial length in a certain direction, represented by the coordinate X (distance from the origin), which has the dimension of length. So far we haven't specified any units. Now we decide to use the length L of King George's foot as our unit of length. This means we divide every value of X by the length L. Letting x denote the numerical value of X in units of feet, we have x = X/L.

Bear in mind that neither X nor L has intrinsic units, and they both have the dimension of length. When we take the ratio of those two lengths, we get a pure dimensionless number, but then we DEFINE this ratio x to be the numerical value of X in units of feet. Students get confused because the ratio of any length to the length of King George's foot is clearly dimensionless, so how can we say the ratio has the dimension of length?

The answer is that we DEFINE the ratio of any length X to the reference standard length L as the magnitude of the length X in units of L. The reason this is confusing is because if we divide X by any other length Y, we usually regard the ratio as just a dimensionless number. However, we COULD regard it as the value of the length X in units of Y, if we were treating Y as simply an arbitrary reference length. The point is that reference lengths defining units are treated (semantically) differently - by convention - from other lengths.

Getting back to the topic of the discussion, when you work with the quantity t/T where T is a reference constant, you are just establishing the units (reference lap times), and the dimensionless ratio gives the numerical value of the coordinate time (which has dimension of time) in units of "reference lap times".

You really seem to honestly believe that a dimensionless number nonetheless has dimensions of time.

To be more precise, the numerical value, in specified units, of any quantity with a certain dimension is always a dimensionless ratio of the magnitude of the quantity to the magnitude of whatever reference quantity defines the units.

Again, this all just confirms that you just integrated dt/gamma.

 

[Austin0]

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).

Thank you for your explication but as I fully understood the meaning of the interval and the metric I am afraid it completely missed the point.
That point being the meaning of the word invariant. In this context it simply means constant,unchanging, across all inertial coordinate systems
This necessarily implies the existence of other frames.
Would you disagree?

It does not apply to other coordinate systems within a single frame. I.e. changing from orthogonal to polar coordinates for eg.

It does not apply to local measurements as they apply within the frame.

It takes local coordinate measurements and outputs a value that is meaningful and constant in all other frames.

As that output value is related to the input values by the gamma factor it would appear it was a de facto transformation, semantic quibbles notwithstanding.
Yes

 

[Austin0]

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: . 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.

"the gamma relationship of space and time". is the inherent relationship that as the spatial part of motion increases relative to the temporal component, the temporal part is decreased relative to other inertial frames. It appears that it was this relationship that determined the metric signiture, the only deviation from a Euclidean metric.
would you agree that this seems to be an intrinsic quality of spacetime geometry??
.

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 dxi, not just dt, and in fact in some metrics there is no dt to begin with.

This time I was presenting an analogy ;-) in this case the black box represented whatever math you cared to employ to achieve your result. that it also entailed a dx was overlooked for simplicity.

The point of course being what was inside the box was irrelevant , The outcome making it clear that in some form the gamma factor had to be present in the box and that the operation itself was functionally a gamma (Lorentz) transformation what ever it might be called.

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.

Yes you did accomplish your goal without explicitly invoking gamma or doing a transformation within a limited semantic loophole regarding what is called a transformation.

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}

Looking at your math it appears to me that at this point =a\sqrt{T^2-4\pi^2 R^2}

you have a distance measurement and a time interval. That you are simply applying the Pythagorean operation to these quantities. the operation itself composites vector components and returns a single value of length. in this case a spacetime length expressed in units of proper time.
The fact that the spatial measurement was not linear does not seem relevant as by this point it is simply a value.
Harking back to an earlier proposition of mine that the gamma was hiding in the pythagorean operation as a geometric gamma function , this seems to be supported by your calculations.
As you say, you did not explicitly employ the function or a gamma value , yet the end result was in fact, transformed.
Since what was in the box was the pythagorean operation this would imply that it was the culprit.
This transformation can be easily demonstrated regarding linear motion within the context of Minkowski geometry.
Regarding the other's arguments about integrating with gamma. In this case with motion limited to angular displacement , is integration even necessary?.
wouldn't it be a simple measurement of distance determined by pi?

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.

T

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

Wouldn't it be more correct to say it was true in general but not necessarily true in certain limited cases like rotating frames , which I would think was not even an inertial frame. This is SR

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.

I agree, cumbersome ;-)
yes I understand the spacetime interval,,, the question was; Was determining the interval a transformation.?

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.
Functionally equivalent to absolute time.

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

We are talking about SR , inertial frames, Not GR or accelerating frames which are outside the topic of this thread and discussion.
No offense intended whatsoever, but you have a general habit of taking questions and statements that have an explicitly limited context and subject, and either answering them with generalities or refuting them for lack of generality.

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.

Wouldn't it be truer to say it is applied within every inertial frame regarding every "other" inertial frame.
In any Minkowski diagram the designated rest frame is in a purely Cartesian coordinate structure with a Euclidean metric or do you disagree with this??
If you disagree could you explain what you mean by applying to every frame?

 

[universal_101]

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.

Thanks for the support on the basic math magic trick, the same can be criticized by the Dalespam's own analogy of Principle of least action and Newtonian mechanics, or even the simple following basic mathematics principle,

which is A+A = 2A, that is, suggesting A+A is not same as 2A, because he never uses multiplication when he uses addition and vice-versa, is entirely incorrect.

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".

Physics implies causality principle, i.e. If we have an effect then we must have a cause, atleast in the domain of classical physics.

Therefore, the effect(Time Dilation of unstable particles) must have a cause(where it comes from). Since, atleast you are accepting that we do have to involve gamma factor in order to understand these effects. How do you understand, the use of this gamma factor which is a part of the transformations, to produce the effects is question.

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.

As I explained, you simply misunderstood what I was trying to ask, and not for introducing personal war of words, but it does not interests me a bit, how easily you make your perspective of other people, just by reading posts from a forum.

 

[universal_101]

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.

In Minkowski spacetime, physical laws are unaffected(i.e.produce invariant results), aren't they ?

Then how come we explain, the invariant results of Muons Experiment, by using a property of Minkowski spacetime ?

Whereas, it seems evident(using basic physical concepts) that the explanation must come from a physical law !

 

[universal_101]

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

This is a good description, but what I'm questioning is the simple incompatibility of the two.

That is, the transformation cannot produce a function which can be applied to physical effects, nor can the physical effects be associated with the transformations. This is the basic physics argument.

And Since, we don't observe any other such physical effects(Length contraction, increased density), just as the above argument suggests.

This implies, we must have a physical law to account for the Time Dilation of unstable particles and increase in inertia, whereas, EM phenomena comes validly under Lorentz transformations.

 

[universal_101]

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

Thanks Herald, I appreciate the insights or answer.

Unfortunately it is incomplete, even if I consider that stretch, Since Special Relativity is all about the transformations of one electromagnetic effect, to be analysed from the other frame so that equations give same result in this another frame too. Special relativity is not about the electromagnetism but it's transformation.

 

[harrylin]

[..]the transformation cannot produce a function which can be applied to physical effects, nor can the physical effects be associated with the transformations. This is the basic physics argument.

Argument of what? Apparently you missed my post #95 (as well as #120). Already in classical mechanics are physical effects associated with the transformations*; thus your denial is simply wrong. Physical laws result in the validity of certain transformation equations between reference systems, so that they are related. If you know the effect (the transformation equations) then you can draw some conclusions about physical laws and causes.

*PS I had in mind Newton's law but did not elaborate, however I see that now there is a fresh thread on that (simple explanation in post #10):
https://www.physicsforums.com/showthread.php?t=610258
Do you really claim that this is wrong?

[..] Special Relativity is all about the transformations of one electromagnetic effect, to be analysed from the other frame so that equations give same result in this another frame too. Special relativity is not about the electromagnetism but it's transformation.

It's about both the physical laws and the resulting transformation equations (emphasis mine):

"Poincaré has objected to the existing theory of electric and optical phenomena in moving bodies that [..] the introduction of a new hypothesis [will be] required [..] each time new facts [are] brought to light. Surely, this course of inventing special hypothesis for each new experimental result is somewhat artificial. It would be more satisfactory, if it were possible to show, by means of certain fundamental assumptions [..] that many electromagnetic actions are entirely independent of the motion of the system. [..] I believe now to be able to treat the subject with a better result. The only restriction as regards the velocity will be that it be smaller than that of light."
- Lorentz 1904

"It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. [..] [SR is] a simple and consistent theory of the electrodynamics of moving bodies based on Maxwell's theory for stationary bodies."
- Einstein 1905

 

[stevendaryl]

Thank you for your explication but as I fully understood the meaning of the interval and the metric I am afraid it completely missed the point.
That point being the meaning of the word invariant. In this context it simply means constant, unchanging, across all inertial coordinate systems. This necessarily implies the existence of other frames. Would you disagree?

Yes, I would disagree. The length of the hypotenuse of a triangle is, by Pythagorus, equal to √(A² + B²), where A is the length of one leg, and B is the length of the other leg. Do you think that that definition implies the existence of other frames? The invariant interval in SR, τ = √((ct)² - x²) is a geometric relationship, just like the length of the hypotenuse of a right triangle. It doesn't have anything to do with "frames". The metric doesn't have anything to do with frames. Curved surfaces have associated metrics, and that doesn't have anything to do with frames.

 

[stevendaryl]

It does not apply to other coordinate systems within a single frame. I.e. changing from orthogonal to polar coordinates for eg.

Yes, it certainly does. In cartesian coordinates, the metric tensor is defined by
(in 2-D spacetime):

g_tt = 1
g_xx = -1/c²
g_yy = -1/c²

(all other components are zero)

In polar coordinates, we have:
g_tt = 1
g_rr = -1/c²
g_θθ = -r²/c²

So the invariant interval is

ds² = dt² - dr²/c² - r²dθ²/c²