Gravitational redshift and length contraction factors

[swansont]

Can you look at the post just above yours where it shows how to calculate the photon travel time along a geodesic?

Yes

[math]\frac{dt}{dr}=\frac{1}{c(1-r_s/r)}[/math]

Unless you're claiming that r_s is the same for the earth and the moon, the time will be different.

[md65536]

[math]\frac{d\tau }{dr}=\frac{1}{c \sqrt{1-r_s/r}}[/math]

Does this give you the change in rate of proper time of a clock as it changes distance from the gravitational mass? No wait... I see, the change in proper time per change in distance. Is this for increasing radius only?

 

 

 

By the way I haven't downvoted a single reply in this thread. I haven't even understood one. Please stop libelling me.

 

 

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I think I understand now.

 

If you took a ruler and a clock from A to B, and you measured the time it takes light to cross the ruler, then added it all up... This would be the equivalent of laying N rulers end-to-end, each with its own clock, and each measuring the time that light takes to cross the 1 light-unit ruler as 1/c units of time.

 

It would be the same if you took a ruler and clock from A (at each step the ruler/clock would match the local ruler/clock) or from B.

 

If this is what is called "ruler distance" then it is the same whether measured from A or from B. This might be the (invariant) "proper length" between an event at A and B.

 

 

Radar distance, as measured from A or B without transporting the clock, doesn't have to agree. A's clock will tick at a different rate relative to any of the N clocks. Due to time dilation between A and B, it must be that the radar return-trip time calculated using only a local clock, must differ between A and B. If converted into the time of a transported clock, then those converted times will be equal.

 

So radar times must be different, but radar distance would depend on how distance is calculated. Does it just use v=c?, or does it account for differences all along the length of rulers? I'm assuming it just uses v=c, otherwise it would probably need to know the ruler distance in order to compensate for it...

 

 

Anyway this much seems certain: The distance and time measured by a transported ruler and clock would be the same, measuring from A to B or vice versa. The return-trip time of a radar signal would be different, measured from A verses from B.

Edited May 25, 2013 by md65536

[xyzt]

Yes

[math]\frac{dt}{dr}=\frac{1}{c(1-r_s/r)}[/math]

Unless you're claiming that r_s is the same for the earth and the moon, the time will be different.

Yes, I already showed that [math]\tau_{ME} \ne \tau_{EM}[/math]

 

Radar is a ROUND TRIP signal:

-one leg from Earth to Moon

-the other leg back from Moon to Earth

 

In the reverse direction:

 

-one leg from Moon to Earth

-the other leg back from Earth to Moon

 

so:

 

[math]\tau_{EME}=\tau_{ME}+\tau_{EM}=\tau_{EM}+\tau_{ME}=\tau_{MEM}[/math]

 

 

 

 

So radar times must be different, but radar distance would depend on how distance is calculated. Does it just use v=c?, or does it account for differences all along the length of rulers? I'm assuming it just uses v=c, otherwise it would probably need to know the ruler distance in order to compensate for it...

Incorrect. Radar is a ROUND-TRIP signal, so, the time Earth-Moon-Earth is equal to the time Moon-Earth-Moon:

 

[math]\tau_{EME}=\tau_{ME}+\tau_{EM}=\tau_{EM}+\tau_{ME}=\tau_{MEM}[/math]

 

So, if you aren't negging the posts that are explaining how things work, who is the coward who's doing that?

Edited May 25, 2013 by xyzt

[md65536]

Incorrect. Radar is a ROUND-TRIP signal, so, the time Earth-Moon-Earth is equal to the time Moon-Earth-Moon:

Do you understand that we're talking about 2 different clocks here, one at A and one at B? Do you understand that these two clocks do not pass through the same pairs of events? Do you understand that the proper time of 2 events at A is generally different from the proper time between two *different* events at B?

 

Am I not understanding radar time correctly? A radar operator sends out a signal, and times the duration until receiving the reflected signal, using a clock that remains stationary relative to the observer... is that not what radar time is?

[xyzt]

Do you understand that we're talking about 2 different clocks here, one at A and one at B?

Yes, I understand. Unfortunately it is clear that you do not understand how time is being measured. You do not understand this for either flat spacetime (see the thread on the twins paradox) and you understand it even less for the case of curved spacetime (see this thread). That's too bad.

[md65536]

Now, if you want the proper time, you need to use the fact that coordinate time [math]dt[/math] is related to proper time [math]d\tau[/math] via the relation:

 

[math]d\tau=dt \sqrt{1-r_s/r}[/math]

You've got it right here! Letting Tau be the proper time measured by your probe, let tA be the coordinate time measured at A, with distance rA. Let tB be the coordinate time measured at B, with a different rB. Then tA and tB are not the same.

[xyzt]

You've got it right here! Letting Tau be the proper time measured by your probe, let tA be the coordinate time measured at A, with distance rA. Let tB be the coordinate time measured at B, with a different rB. Then tA and tB are not the same.

 

 

Sigh,

 

[math]\tau_{EM}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sE}/r}}}[/math]

[math]\tau_{ME}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sM}/r}}}[/math]

 

[math]\tau_{EME}=\tau_{ME}+\tau_{EM}=\tau_{EM}+\tau_{ME}=\tau_{MEM}[/math]

Edited May 25, 2013 by xyzt

[md65536]

Unfortunately it is clear that you do not understand how time is being measured.

With a clock. There's a clock at A. There's a different clock at B. There's another clock on your probe (depending).

[xyzt]

With a clock. There's a clock at A. There's a different clock at B. There's another clock on your probe (depending).

Doesn't matter, you are fixated on the fact that the clocks are placed on different celestial bodies (Earth vs. Moon). Yet GR, teaches you how to do the calculations correctly, so the radar times , despite your nonsense, are the same for both the EME and the MEM trips. Try taking a class, it will cure you from making all these nonsense statements that are clearly incorrect.

 

You've got it right here! Letting Tau be the proper time measured by your probe, let tA be the coordinate time measured at A, with distance rA. Let tB be the coordinate time measured at B, with a different rB. Then tA and tB are not the same.

This is total gibberish, sorry, I just showed you the correct mathematical solution.

Edited May 25, 2013 by xyzt

[md65536]

[math]\tau_{EM}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sE}/r}}}[/math]

[math]\tau_{ME}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sM}/r}}}[/math]

 

[math]\tau_{EME}=\tau_{ME}+\tau_{EM}=\tau_{EM}+\tau_{ME}=\tau_{MEM}[/math]

Now try calculating the proper time according to a clock that stays on Earth.

[swansont]

Yes, I already showed that [math]\tau_{ME} \ne \tau_{EM}[/math]

 

 

Sigh,

 

[math]\tau_{EM}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sE}/r}}}[/math]

[math]\tau_{ME}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_{sM}/r}}}[/math]

 

You said earlier that [math]\tau[/math] is an invariant.

[xyzt]

Now try calculating the proper time according to a clock that stays on Earth.

You don't get , do you? The signal climbs out of the gravitational well of the Earth*, so, BOTH the Earth - bound clock and the Moon bound clock AGREE that it takes the PROPER time [math]\tau_{EM}[/math] for the signal to reach the Moon. Actually, all the observers in the universe agree on [math]\tau_{EM}[/math].

In the reverse direction: The signal climbs out of the gravitational well of the Moon, so, BOTH the Earth - bound clock and the Moon bound clock AGREE that it takes the PROPER time [math]\tau_{ME}[/math] for the signal to reach the Earth. [math]\tau_{ME}[/math] need not be equal to [math]\tau_{EM}[/math].

You need to learn the meaning of proper time, all clocks agree on the elapsed proper time.

You said earlier that [math]\tau[/math] is an invariant.

These are two DIFFERENT directions of the signal, why would you expect them to be the same proper time? Do you understand the notion of invariance? (Hint: I just explained it above).

 

 

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* Things are a little more complicated than that , the signal climbs out of the gravity well of the Earth and falls into the gravity well of the Moon (and vice-versa). This does not affect the symmetry of the problem.

Edited May 25, 2013 by xyzt

[swansont]

Yes, I do. Time is a scalar, though.

[xyzt]

Yes, I do. Time is a scalar, though.

Based on your question, you don't. [math]\tau_{EM}[/math] is the same for all the observers in the universe. [math]\tau_{ME}[/math] is the same for all the observers in the universe.

The above does neither require nor imply that [math]\tau_{EM}=\tau_{ME}[/math] . Isotropy would require that but the setup is not isotropic, the path from Moon to Earth is not identical with the path from Earth to Moon.

You can make an argument that both Schwarzschild radii [math]r_{sM},r_{sE}[/math] are so small that the two proper times are very close to each other (and very close to the Newtonian value [math]\frac{D}{c}[/math] as I explained early on.

On the other hand, in a setup where you would have celestial bodies with a very large gravitational mass, one cannot ignore the Schwarzschild radii and, if there is a big disparity between the Schwarzschild radii, the two proper times will be significantly different and the approximation [math]\frac{D}{c}[/math] will no longer hold. To my best recollection, the EFEs have a solution (Schwarzschild) only for the case when the mass of one celestial body is much larger than the second body.

Edited May 26, 2013 by xyzt

[md65536]

You need to learn the meaning of proper time, all clocks agree on the elapsed proper time.

I was under the impression that different clocks measure generally different proper times.

[xyzt]

I was under the impression that different clocks measure generally different proper times.

For a unique path through spacetime? No. Quite the opposite.

Edited May 26, 2013 by xyzt

[md65536]

For a unique path through spacetime? No. Quite the opposite.

Do clocks in different locations follow a common path through spacetime?

[xyzt]

Do clocks in different locations follow a common path through spacetime?

If they are comoving, yes.

[DimaMazin]

Based on your question, you don't. [math]\tau_{EM}[/math] is the same for all the observers in the universe. [math]\tau_{ME}[/math] is the same for all the observers in the universe.

The above does neither require nor imply that [math]\tau_{EM}=\tau_{ME}[/math] . Isotropy would require that but the setup is not isotropic, the path from Moon to Earth is not identical with the path from Earth to Moon.

What is shorter?

[md65536]

If they are comoving, yes.

Okay, some of the things you've said in other threads are now starting to make sense to me. So if you have 2 events along a timelike path P and tau is the proper time between them according to some clock that follows P, any clock that is comoving with it follows the same path and passes through the events also? Thus not only agreeing on the proper time of the separated clock but also measuring it too. All along I thought that events at A were not passed through by (comoving) observer B.

[DimaMazin]

Okay, some of the things you've said in other threads are now starting to make sense to me. So if you have 2 events along a timelike path P and tau is the proper time between them according to some clock that follows P, any clock that is comoving with it follows the same path and passes through the events also? Thus not only agreeing on the proper time of the separated clock but also measuring it too. All along I thought that events at A were not passed through by (comoving) observer B.

Earth measures photons travel with own clock.Moon measures photons travel with own clock.The measurements aren't equal.

[swansont]

Based on your question, you don't. [math]\tau_{EM}[/math] is the same for all the observers in the universe. [math]\tau_{ME}[/math] is the same for all the observers in the universe.

The above does neither require nor imply that [math]\tau_{EM}=\tau_{ME}[/math] . Isotropy would require that but the setup is not isotropic, the path from Moon to Earth is not identical with the path from Earth to Moon.

OK. That seems to be what was confusing me. (D'oh!)

 

So a one-way signal measurement would yield a different length, since it depends on the source, but not a round-trip measurement.

[xyzt]

 

OK. That seems to be what was confusing me. (D'oh!)

 

So a one-way signal measurement would yield a different length, since it depends on the source, but not a round-trip measurement.

yes, now if the ignorant coward who's been negging my posts on the subject would come forward...

Edited May 26, 2013 by xyzt

[StringJunky]

yes, now if the ignorant coward who's been negging my posts on the subject would come forward...

 

My experience here tells me you are getting negged for negative attitude and delivery, which you are showing in spades, not for appearing to be wrong.

[xyzt]

My experience here tells me you are getting negged for negative attitude and delivery, which you are showing in spades, not for appearing to be wrong.

The delivery is purely scientific, there is someone who's an ignorant coward who's negging the posts.

Edited May 26, 2013 by xyzt