# Observable physical quantity in relativity

1. Oct 22, 2008

### bernhard.rothenstein

Please let me know when do we say that a physical quantity is observable. Is the answer related to clock synchronization?. If possible do not involve quantum physics.

2. Oct 22, 2008

### Makep

What quantity are you thinking about? An event or energy or mass or what?

3. Oct 23, 2008

### bernhard.rothenstein

Thanks for your question which leads to clarification. Say the physical quantities involved in a Lorentz transformation process.

4. Oct 23, 2008

### HallsofIvy

The only place I have ever seen "observable" used is in Quantum Physics!

5. Oct 23, 2008

### bernhard.rothenstein

Thanks
observable=measurable in SRT.

6. Oct 23, 2008

### dx

The only "observables" one normally deals with in SR are the coordinates of events. You can imagine a grid of clocks arranged like a lattice in space, and synchronized according to Einstein simultaneity. Then, these clocks can record the times of events taking place in their immediate vicinity. This determines the standard (x,y,z,t) coordinates of the event. Different synchronization procedures give different coordinatizations of spacetime.

But it is not necessary to use such a grid, since you can characterize events with purely local measurements of time made by the observer, i.e the times between various events on the observers world line.

All other physical quantities that we consider observable in classical physics are functions of the coordinates of various events.

Last edited: Oct 23, 2008
7. Oct 23, 2008

### DrGreg

Bernhard, knowing the sort of papers you read, I think I know what you are asking.

(In quantum theory, "observable" has a technical meaning, but here we are interested in classical, i.e. non-quantum, theory.)

There is a distinction to be made between observable quantities that you can unambiguously measure without any knowledge of physics, and other quantities which you calculate from observables, the calculation involving applying some theory of physics.

For example, proper time, measured by a clock carried by an observer between two events that both occur at the same location as the observer, is an "observable" quantity. So too is the frequency of a wave at the location of the observer, also measured by the observer's own clock. In the old days when we defined length using a ruler, the rest length of an object stationary relative to an observer was an observable quantity.

But coordinate time is not an observable because it involves an application of Einstein's synchronisation convention (or some other convention). (Thus I disagree with dx, post #6.) In other words, coordinate time depends on which theory or convention you choose to use. And therefore derived quantities such as coordinate velocity and coordinate acceleration aren't observable either. However proper acceleration of the observer is observable because you can measure it directly using an accelerometer.

(Note that it is unobservable coordinate time that appears in the Lorentz transform (except when x=0 and then coordinate time = proper time). Note also that the length of an object moving relative to an observer is not observable, either, because it requires two positions to be recorded simultaneously, which depends on your choice of simultaneity.)

It is arguable whether the modern definition of distance-from-observer using radar (and an inertial observer) is an observable or not. You could argue that it isn't because it involves the constancy of the speed of light, a theoretical assumption. However you could also argue that the constancy of the speed of light isn't theory at all, simply a logical consequence of our current definition of distance, so I think you could say radar distance actually is an observable. (The theory comes in when we postulate that radar distance and ruler distance are equal.)

In the same spirit, I think you could perhaps argue that doppler shift (the k of k-calculus) is also an observable, given that you can calculate it directly from two observables, emitter frequency and receiver frequency, and the calculation is an irrefutable definition rather than a potentially refutable theory. However I'm not entirely convinced by this argument.

8. Oct 23, 2008

### bernhard.rothenstein

Thanks DrGreg. You guessed where from my problem arises.
Considering the transformation equations that result from the fact that in one of the involved inertial reference frames the clocks are standard synchronized whereas in the other the clocks are “everyday” synchronized (we have discussed it on the Forum) by a signal that propagates apparently with an infinite speed, they have the shape
x=x’sqrt[(1+V/c)/(1-V/c)]+gVt’ (1)
T=gt’+[(1+V/c)/(1-V/c)]x’/c (2)
the inverse ones being
x’=g(x-VT) (3)
t’=sqrt[(1+V/c)/(1-V/c)]t (4)
T time displayed by standard synchronized clocks, t and t' times displayed by everyday synchronized clocks.
In accordance with what I have learned from your answer, the lengths measured using the radar detection procedure, i.e. related by the Doppler factor are measurable (first term in the right side of (1) second term in the right side of (2) and so are the time intervals related by the Doppler factor as in (4) are measurable.
What about the terms related by the Lorentz factor g.
Could we say that following Reichenbach’s relativity theory and deriving relativistic formulas which do not contain his anisotropy factor, they relate measurable physical quantities? Is that the situation in the case of (3) which has the same shape in all “theories” that correspond to different values of the anisotropy factor?

9. Oct 23, 2008

### dx

DrGreg,

With your definition of observable, the only observable in any theory of dynamics would be proper time. You say that frequency is an observable because it is a function of proper times between events near the observers own clock, but notice that even coordinate distance and coordinate time can be defined in this way. There are other ways to define them, but even frequency can be defined in other ways, and the only justification for any particular way to define it comes from theory.

10. Oct 24, 2008

### DrGreg

On further thought I would be reluctant to describe as "observable" or "measurable" something that you calculate from more than one quantity you have measured. I think I'd reserve that name for raw measurements.

In your case, quantities that are independent of Reichenbach's $\epsilon$ factor could be described as "independent of synchronisation" or "sychronisation-free" or something like that. Ultimately, whatever terminology you use, you need to explain it clearly to your reader.

11. Oct 25, 2008

### bernhard.rothenstein

Consider the "radar echo" experiment in which a stationary source emits at a time t(e) a light signal toward a moving mirror which reflects it back being received at the location of the source at a time t(r). If the mirror moves with speed V and the time intervals dt(e) and dt(r) are measured using the same clock then
dt(r)=dt(e)[(1+V/c)/(1-V/c) (1)
The derivation of (1) is clock synchronization free.
There are other effects which relate the same physical quantities via the square root of [(1+V/c)/(1-V/c)] (Doppler shift, Radar detection of moving rods). Is it correct to say that they are synchronization free (measurable).
Thanks for your answer.

12. Oct 25, 2008

### MeJennifer

Practically speaking yes.

But theoretically in might be interesting to consider the momentum of the photon reaching the mirror and the time and space between the absorbtion and emittance of the photon at the mirror which are frame dependent due to resp. the acceleration and the relative speed between the stationary source and the mirror.

13. Oct 27, 2008

### DrGreg

dte and dtr are "measurable" but V is not. V is coordinate velocity, i.e. coordinate distance divided by coordinate time. Coordinate time certainly does depend on synchronisation, so any formula involving V is synchronisation-dependent.

Note however that the term

$$\sqrt{\frac{1 + V/c}{1 - V/c}}$$​

is the doppler factor which is synchronisation-independent and can be easily calculated from two "measurable" frequencies. So if you use the symbol k (e.g.) for this quantity and derive the relevant equations without reference to V, then you have a synchronisation-independent argument.

If you want to completely avoid synchronisation, instead of saying A moves relative to B "with velocity V" you can instead say "...with doppler shift k".

Note also that "rapidity" is just logek (or c loge k if you prefer it in units of velocity) and therefore also a synchronisation-independent quantity, so you could also say "...with rapidity $\phi$".

14. Oct 29, 2008

### bernhard.rothenstein

Thank you.
1. Please tell me which is the best name for epsilon dependent transformation equations.
2. All theories that result from different values of the synchrony parameter epsilon lead to the same transformation equation for the space coordinates
x'=g(x-Vt)
which is epsilon independent. Is there an explanation for that. Is that the result of the fact that it is derived considering length contraction and time dilation two relativistic effects that compensate each other??

15. Oct 29, 2008

### DrGreg

1. I'm not aware of any official name. I'd just say "synchronisation-dependent" or maybe even "convention-dependent". I think maybe some authors might say "conventional" but I think that's a confusing word to use in this context.

2. This is because, in your scenario, distance x' is defined in a sync-independent way e.g. as radar distance which depends only on two proper-time measurements and the 2-way speed of light but not the one-way speed of light. It takes the same value no matter what you choose for $\epsilon$. Whereas x, V and t (and $\gamma$) are all defined for a single coordinate system only (your "reference" Einstein-synced system) and do not vary.

Note that x' is the distance between an event and an observer; it is not the distance between two events.

16. Nov 1, 2008

### bernhard.rothenstein

Re. Observed and reckoned physical quantities

If we use the formula that accounts for the radar echo (police radar) in order to reckoned the speed V we speek about a calulated physical quantity? Does special relativity trust in calculated ohysical quantities?

17. Nov 3, 2008

### DrGreg

Re: Re. Observed and reckoned physical quantities

Coordinate velocity V = dx/dt is synchronisation-dependent because coordinate time t is synchronisation-dependent. (If you like, V is a one-way speed, not a two-way speed; you already know that the one-way speed of light can differ from the two-way speed in non-standard coordinates.)

In special relativity, we almost always assume that Einstein's synchronisation convention applies, so there's no need to worry about whether something is synchronisation-dependent or not. In fact, most people are not aware of the issue, and that is not a problem because it doesn't matter.

I'm not sure I'd like to say velocity derived from doppler is a "calculated physical quantity": it's a "conventional" quantity (by which I mean "depending on a choice of convention"), and so in coordinates with a different Reichenbach sync parameter you would get a different velocity for the same measured doppler shift.

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