Adjusting length and period using (inverse) transformations

[ANvH]

Can you find any reference that defines Doppler in terms of your "non-standard position"?

Yes I do: http://philsci-archive.pitt.edu/1743/2/Norton.pdf
The author uses the Lorentz transform on a waveform. I think he wrote an excellent article about Einstein and Lorentz. The author derived the Doppler shift the way I had presented in post #16. It did not involve the concept of time dilation and length contraction. To better understand time dilation and length contraction I came here, trying to reconcile this with my "non-standard position" (which initially sounded a bit offensive). If this reference I quote above belongs to something "non-standard" then I hope you would read up, because the author hails the Lorentz transformations and hails Einstein.

If you want to change the scenario every time you transform to a new IRF by eliminating the original observer and inserting a new observer at rest in the new IRF, then it's perfectly understandable why you get a different Doppler for every IRF. But if you don't eliminate the original observer, then he will have the same Doppler in the new IRF as he did in the IRF in which he is at rest or any other IRF.

I know that the original observer will have the same Doppler in the new IRF. I thought you understood that. I was talking about a new observer and this new observer will see another Doppler. I am very well aware of what invariance means. Hopefully, this will clear up the misunderstanding. I also hope that it is true that when a source is regarded in a IRF at rest, that the observer in motion will measure a Doppler shift.

There's no confusion here now that you have explained what you are doing. If you want to claim that your position is the "standard" one, then you need to provide support for that from a reference, you just can't go changing well-established definitions.

I am claiming that my position is standard, because I am using Lorentz transformations.

 

[ghwellsjr]

http://www.einstein-online.info/spotlights/doppler

That's a very good reference.

I am quoting

"the relativistic Doppler effect is different from its classical counterpart. It takes into account relativistic time dilation. As it turns out, time dilation and the classical Doppler effect combine in precisely the right way to eliminate the difference between the motion of the source and that of the receiver. True to form, the relativistic Doppler effect depends only on the relative motion of source and receiver."

It looks like what you say "single source and receiver". I don't get it when a bunch of observers each having a different motion would measure the same Doppler effect.

They're not going to measure the same Doppler effect. Where'd you get that idea?

Each observer defines their frame of reference, applies the Lorentz transform of their relative speed to the source. I am being told this is not correct.

Did I say it's not correct? It's just not necessary. Each observer just compares the frequency of the incoming signal to his own time reference. He doesn't need to define a frame of reference or know anything about the Lorentz transform or even know what the relative speed to the source is. But if he does know all of that, then he will actually have to send out some radar signals and make observations and log all the information, and after he makes the assumption that the radar signals travel at the same speed away and returning, he can establish the scenario in his own rest frame and then transform to any other frame.

Should I spell out exactly how I carry out a Lorentz transform on a wave, would that shed light?

I don't see how it will shed any light, since you already explained that you assume in each IRF, that there is an observer at rest and so you change the scenario.

 

[ghwellsjr]

Can you find any reference that defines Doppler in terms of your "non-standard position"?

Yes I do: http://philsci-archive.pitt.edu/1743/2/Norton.pdf
The author uses the Lorentz transform on a waveform. I think he wrote an excellent article about Einstein and Lorentz. The author derived the Doppler shift the way I had presented in post #16. It did not involve the concept of time dilation and length contraction. To better understand time dilation and length contraction I came here, trying to reconcile this with my "non-standard position" (which initially sounded a bit offensive). If this reference I quote above belongs to something "non-standard" then I hope you would read up, because the author hails the Lorentz transformations and hails Einstein.

That's a very long paper. Can you identify the page number that is significant to you?

I know that the original observer will have the same Doppler in the new IRF. I thought you understood that.

Mission accomplished. Thanks for acknowledging this.

I was talking about a new observer and this new observer will see another Doppler. I am very well aware of what invariance means. Hopefully, this will clear up the misunderstanding.

Yes, that clears up the misunderstanding.

I also hope that it is true that when a source is regarded in a IRF at rest, that the observer in motion will measure a Doppler shift.

It's true.

I am claiming that my position is standard, because I am using Lorentz transformations.

That doesn't make it standard.

 

[ANvH]

That's a very long paper. Can you identify the page number that is significant to you?

Page 40, Equation (19)

That doesn't make it standard.

What makes it non-standard? The line metric equation

$c^{2}Δ\tau^{2}=c^{2}Δt^{2}-Δr^{2}$​

is preserved by the Lorentz transformation. If we boost a waveform in the x-direction, the Lorentz transform changes the temporal and spatial characteristics of the waveform. By doing so, the speed of light is not affected.

I did not know that applying a boost to a waveform in the way I presented it earlier, aka, the way Norton in his lengthy paper described, is not a main-stream approach. I really thought that you and others were aware that the relativistic Doppler effect follows directly from this transformation.

My assumption apparently was wrong, leading to a number of misunderstandings. So yes, finally, mission almost accomplished.

 

[ghwellsjr]

That's a very long paper. Can you identify the page number that is significant to you?

Page 40, Equation (19)

That equation is showing how the frequency of light from a moving source will be determined at a location that is not moving. It is deriving the same Doppler shift formula that Einstein showed in article 7 of his 1905 paper introducing relativity, except that Einstein did it for a stationary source and a moving observer and he gets the inverse relationship.

Yes, everyone agrees on the relativistic Doppler shift formula (or its inverse, depending on how you define the sign of the relative velocity between the source and the observer).

But that is not what we are talking about when we transform both the source and the observer in relative motion into any other arbitrary IRF. The Doppler that the observer sees remains the same.

That doesn't make it standard.

What makes it non-standard? The line metric equation

$c^{2}Δ\tau^{2}=c^{2}Δt^{2}-Δr^{2}$​

is preserved by the Lorentz transformation. If we boost a waveform in the x-direction, the Lorentz transform changes the temporal and spatial characteristics of the waveform. By doing so, the speed of light is not affected.

I did not know that applying a boost to a waveform in the way I presented it earlier, aka, the way Norton in his lengthy paper described, is not a main-stream approach. I really thought that you and others were aware that the relativistic Doppler effect follows directly from this transformation.

My assumption apparently was wrong, leading to a number of misunderstandings. So yes, finally, mission almost accomplished.

You are saying that in one particular IRF, the waveform in transit between a source and observer has certain characteristics and in another IRF it has different characteristics, and if you make it clear that that is what you are talking about when you use the term "Doppler", then no one will object. The problem came about when you denied that the observer would see the same Doppler in this new IRF than what he saw in the original IRF. But that has been cleared up, so all the misunderstandings have been resolved.

 

[ANvH]

so all the misunderstandings have been resolved.

I think it is. I'd like to have more insight with respect to measuring proper time.

If the definition of proper time (as distinguished from coordinate time) is unfamiliar, I'd suggest looking at http://www.eftaylor.com/special.html, which has the first few chapters of an older edition of a standard SR textbook online, "Spacetime physics". Actually, I'd suggest reading it anyway :-)

I'll go with what pervect said. I want to make sure. Given

$c^{2}Δ\tau^{2}=c^{2}Δt^{2}-Δr^{2} (1)$​

where $Δ\tau$ is the proper time lapse and $Δt$ is the coordinate time lapse, and working out this equation further,

$Δt=\gamma Δ\tau.$​

This equation suggests that $Δ\tau$ is smaller than the coordinate time lapse, yet you showed by the diagrams that the proper time lapse is increased. So may be I am still confused.

With respect to the proper distance and coordinate distance, using

$Δ\sigma^{2} = Δr^{2}-c^{2}Δt^{2} (2)$​

which looks quite similar to equation (1), as if $Δ\sigma^{2} =-c^{2}Δ\tau^{2}$, but I guess that the minus sign relates to the space-like convention (−+++) used in equation (2). I am arriving at

$cΔt=\gamma Δ\sigma$​

and think I should equate $cΔt$ with the coordinate distance. If I do that then it tells me that the proper distance is smaller than the coordinate distance, which constitutes a length contraction, you also showed by the diagrams.

This brings me all the way back to the start of this thread.

 

[ghwellsjr]

so all the misunderstandings have been resolved.

I think it is. I'd like to have more insight with respect to measuring proper time.

If the definition of proper time (as distinguished from coordinate time) is unfamiliar, I'd suggest looking at http://www.eftaylor.com/special.html, which has the first few chapters of an older edition of a standard SR textbook online, "Spacetime physics". Actually, I'd suggest reading it anyway :-)[/QUOTE
I'll go with what pervect said. I want to make sure. Given

$c^{2}Δ\tau^{2}=c^{2}Δt^{2}-Δr^{2} (1)$​

where $Δ\tau$ is the proper time lapse and $Δt$ is the coordinate time lapse, and working out this equation further,

$Δt=\gamma Δ\tau.$​

This equation suggests that $Δ\tau$ is smaller than the coordinate time lapse, yet you showed by the diagrams that the proper time lapse is increased. So may be I am still confused.

I explained the relationship between Proper Time for a moving clock, Coordinate Time and gamma in my very first post to you (#2):

Time Dilation refers to the delta Coordinate Time of an object compared to its delta Proper Time for the same pair of events. The delta Proper Time is taken between two events for an object in its rest frame. Then you can transform the Coordinate Times for those same two events into a moving frame and get a new delta Coordinate Time. The ratio of the delta Coordinate Time to the Proper Time is the Time Dilation factor and is equal to gamma.

And then I showed a spacetime diagram:

And gave this explanation:

For Time Dilation, we can see that the period from the first event along the blue end of the object to the last event in the moving frame is 5 nanoseconds whereas it was 4 nanoseconds in the rest frame for a ratio of 1.25, the same as the gamma factor.

When we say the Proper Time is dilated, we mean the time interval is stretched out compared to the Coordinate Time. So in the above example, 4 nanoseconds of Proper Time on a clock is stretched out to 5 nanoseconds of Coordinate Time in the frame in which the clock is moving at 0.6c. In other words, 5=1.25*4. Isn't that clear?

With respect to the proper distance and coordinate distance, using

$Δ\sigma^{2} = Δr^{2}-c^{2}Δt^{2} (2)$​

which looks quite similar to equation (1), as if $Δ\sigma^{2} =-c^{2}Δ\tau^{2}$, but I guess that the minus sign relates to the space-like convention (−+++) used in equation (2). I am arriving at

$cΔt=\gamma Δ\sigma$​

and think I should equate $cΔt$ with the coordinate distance. If I do that then it tells me that the proper distance is smaller than the coordinate distance, which constitutes a length contraction, you also showed by the diagrams.

This brings me all the way back to the start of this thread.

Yes, and also to post #2 where I also explained Length Contraction.

Your equations (1) and (2) are different forms of the spacetime interval. It's usually kept as a squared function so as not to deal with the issue of taking a negative squareroot. Having two forms of the equation is another way to deal with the same issue. You use whichever one takes a positive squareroot.

The point of the spacetime interval is that it is the same in all inertial reference frames for any two events. You can work this out for pair of events for the three different frames for the same scenario as shown in posts #11 and #13.

The spacetime interval will fall into one of three categories:

(1) Time-like: the interval can be measured by an inertial clock present at both events as simply the Proper Time interval between those two events. Note that this is the same as transforming to the rest frame of that inertial clock where the Proper Time interval equals the Coordinate Time interval.

(2) Space-like: the interval can be measured by an inertial ruler present at both events and at the same time as simple the spatial difference between those two events. Note that this is the same as transforming to the rest frame of the inertial ruler where the Proper Length interval equals the Coordinate Distance interval.

(3) Light-like: the interval is zero meaning it cannot be measured. This is also called a null-interval. Only light can be present at both events.

The spacetime interval is never "spacetime", it is always either "space" or "time" or neither ("null").

 

[ANvH]

The spacetime interval is never "spacetime", it is always either "space" or "time" or neither ("null").

I understood/understand this as I understood/understand the diagrams. I also realize that it seems to be impossible to make an assertion that there is a discrepancy. Your diagrams, together with explanations, indicate that

$\gamma = \frac{Δt}{Δ\tau}$​

which also follows from the space-time interval (equation (1), my previous post), which says that the coordinate time, $Δt$, is dilated by the factor $\gamma$. This can be easily shown by your diagram(s). Again, I do not disagree, and this describes the space-time interval as time.

You also nicely explained with the same diagram that the coordinate distance between the two ends of a rod is shortened by the factor $\gamma$. The discrepancy I note, using above equation,

$\gamma = \frac{cΔt}{cΔ\tau}=\frac{Δr}{Δ\sigma}$​

suggests that the coordinate distance, $Δr$, is dilated by the factor $\gamma$.

 

[ANvH]

The discrepancy I note, using above equation,

$\gamma = \frac{cΔt}{cΔ\tau}=\frac{Δr}{Δ\sigma}$​

suggests that the coordinate distance, $Δr$, is dilated by the factor $\gamma$.

I guess that since the square of $Δ\sigma$ is negative, the above is irrelevant.