Gravitational redshift and length contraction factors

[SamBridge]

Wouldn't the same faulty reasoning apply to SR? If there's time dilation in SR, wouldn't you predict expanded distances?

If we accept the notion that light is constant, we must accept the notion that the rate time passes and relative distance can change. In Lorentz transformations, time dilation alone nor length contraction alone counts for the whole picture of relative motion to inertial frames when one object is stationary to an outside frame. If you have two objects traveling 99% the speed of light directly away from each other, since they travel at the same speed, their clocks would agree, which can only leave the length to contract because if it didn't they would measure each other traveling 1.98 times the speed of light. It is the same principal as if they were heading towards each other, but because they are heading towards each other, if length contacted it would decrease the distance they see between each other on top of the already classical notion that they would measure each other heading towards each other at 1.98 times the speed of light. So, the length must increase. Didn't you say in your first post a meter stick can get longer?

 

 

 

Also, B's meter stick appears longer than a meter to A,

Edited May 23, 2013 by SamBridge

[md65536]

[math]\tau[/math] is proper time, this is an invariant, remember? we went over this in the other thread, the one where you claiming that acceleration is not important in the twins paradox.

Where did you learn all of this? You definitely have a wealth of knowledge, but I question your ability to apply it (yes, I remember the other thread).

 

So can radar time be measured by radar, ie. light?, and what would the proper time be then?

[xyzt]

Where did you learn all of this? You definitely have a wealth of knowledge, but I question your ability to apply it (yes, I remember the other thread).

 

So can radar time be measured by radar, ie. light?, and what would the proper time be then?

I studied it.

The times are the times I already posted, and yes, they apply to light (or em waves).

I couldn't care less about your "questioning my abilities" because you obviously do not know these subjects. What I don't understand is how can you make pronouncements when you didn't study, can you explain this? You claim to have a BS, in what field is it, if I may ask?

Edited May 23, 2013 by xyzt

[md65536]

I studied it.

Studied where? I'm just curious about how one attains your particular combination of knowledge and abilities.

 

You've done the same thing in this thread as in the other thread you mentioned.

1) Introduce a new, marginally related observer and throw around some calculations for it (eg. the proper time of a traveling particle).

2) Ignore the observations we were discussing, which don't match your calculations, and justify this by claiming that the observations we were discussing are irrelevant to the alternate thing you were calculating.

3) Derail the conversation by focussing entirely on this alternate observer, no longer even bothering to relate it to the original topic. Eg. "The time dilation experienced by two observers with different gravity potential is irrelevant because the proper time of a single observer is invariant."

 

You may be making some true statements, but you're falsely claiming they have any bearing on the topic at hand.

Edited May 23, 2013 by md65536

[xyzt]

Studied where? I'm just curious about how one attains your particular combination of knowledge and abilities.

 

You've done the same thing in this thread as in the other thread you mentioned.

1) Introduce a new, marginally related observer and throw around some calculations for it (eg. the proper time of a traveling particle).

2) Ignore the observations we were discussing, which don't match your calculations, and justify this by claiming that the observations we were discussing are irrelevant to the alternate thing you were calculating.

3) Derail the conversation by focussing entirely on this alternate observer, no longer even bothering to relate it to the original topic. Eg. "The time dilation experienced by two observers with different gravity potential is irrelevant because the proper time of a single observer is invariant."

 

You may be making some true statements, but you're falsely claiming they have any bearing on the topic at hand.

Well, after two threads you still don't understand proper time.

You don't know GR, by your own admission, so how can you pass judgement?

This is not a problem of time dilation, you keep making the same mistake over and over again, this is a problem of elapsed proper time. After two threads and countless corrections, you still don't know the difference.

[md65536]

If A and B are continuously bouncing signals off each other, A might say "each signal takes 1 second round trip to travel, so the distance is 1 light second" and ages 100 seconds per 100 signals, while B might say "each signal takes 1.1 seconds, the distance is 1.1 light seconds," and ages 110 seconds per the same 100 signals.

 

[...]

 

It still doesn't make sense to me, because... say you place metersticks between A and B, say 1000 of them, I would think that each measures a local ruler as 1 meter, and both agree that the rulers at B are bigger than the rulers at A ("rulers shrink in a gravitational field"???). Then A would measure the distance as being greater than 1000 m and B would measure it as less than 1000 m. Am I mistaking the meaning of rulers shrinking in a gravitational field? Is it A who observes a meterstick brought into the field as smaller than a meter, while B observes it as 1 m?

Part of my confusion might come from failing to realize that radar distance and ruler distance are generally different measures of distance. So measuring the timing of light signals and placing rulers end-to-end probably won't give identical results.

 

I think the redshift interpretation above is right, but I'm not so sure about the rulers.

[xyzt]

Part of my confusion might come from failing to realize that radar distance and ruler distance are generally different measures of distance. So measuring the timing of light signals and placing rulers end-to-end probably won't give identical results.

 

I think the redshift interpretation above is right, but I'm not so sure about the rulers.

Physics is about calculating the answers. How would you calculate the answer?

[swansont]

xyzt, on 23 May 2013 - 18:40, said:

Very simple, because the formulas that I have given you already correctly incorporates the effect on gravitational fields on clocks (see the presence of the [math]r_E[/math]?).

Yes, that's what proper time does. But are the observers actually measuring proper time on their clocks? They are not in inertial frames.

[imatfaal]

 

 

Yes, we are talking GR, a subject that I am very familiar with. I can tell you in no uncertain terms that your attempts at calculating are incorrect.

Here is the deal:

A probe particle will :

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

-on the return trip, it will climb out of the gravitational well of the Moon and fall into the gravitational well of the Erath. This will take a time [math]\tau_{ME}[/math]

 

The other probe particle will do the reverse trip:

-climb out of the gravitational well of the Moon and fall into the gravitational well of the Earth. This will take a time [math]\tau_{ME}[/math]

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

 

The total time for either particle (the "radar distance") is the same, [math]\tau_{EM}+\tau_{ME}[/math], just the order of addition is reversed.

 

Now, for people interested in the subject, I could show the the expressions for [math]\tau_{ME}[/math] and [math]\tau_{EM}[/math]. For example:

 

[math]c \tau_{ME}=\sqrt{\frac{r_0}{r_E}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math]

 

 

 

 

 

where [math]r_E[/math] is the Earth Schwarzshild radius, [math]r_0=D[/math] and [math]r[/math] is the Earth radius. The derivation is quite complicated but I can attach a file that shows how it is done.

 

Likewise:

 

[math]c \tau_{EM}=\sqrt{\frac{r_0}{r_M}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math]

 

where [math]r_M[/math] is the Moon Schwarzshild radius.

 

The presence of the Schwarzschild radii encapsulates the gravitational effects on clocks.

 

 

Could you help out with a few figures ?

 

just using rough orders of magnitude

 

[math]c \tau_{ME}=\sqrt{\frac{r_0}{r_E}} \cdot r_0 \cdot arctan \left( \sqrt{\frac{r}{r_0-r}}\right)+\sqrt{r(r_0-r)}[/math]

 

r_0 = 10e8 metres

r_e = 10e-3 metres

r = 10e6 metres

 

one ends up with c tau_me being of the order of 10e13. possibly I have parsed your equation incorrectly - but that figure seems a little screwy and the figure for the moon -earth will be around an order of magnitude higher (schw-rad is a couple of orders of mag smaller)

[xyzt]

Yes, that's what proper time does. But are the observers actually measuring proper time on their clocks? They are not in inertial frames.

The observer floating somewhere between the Earth and the Moon, outside their gravitational fields is clearly inertial.

Could you help out with a few figures ?

 

just using rough orders of magnitude

 

[math]c \tau_{ME}=\sqrt{\frac{r_0}{r_E}} \cdot r_0 \cdot arctan \left( \sqrt{\frac{r}{r_0-r}}\right)+\sqrt{r(r_0-r)}[/math]

 

r_0 = 10e8 metres

r_e = 10e-3 metres

r = 10e6 metres

 

one ends up with c tau_me being of the order of 10e13. possibly I have parsed your equation incorrectly - but that figure seems a little screwy and the figure for the moon -earth will be around an order of magnitude higher (schw-rad is a couple of orders of mag smaller)

The order is [math]\sqrt{10^{10}} \sqrt{10^{14}}=10^{12}[/math].

Edited May 24, 2013 by xyzt

[imatfaal]

The observer floating somewhere between the Earth and the Moon, outside their gravitational fields is clearly inertial.

but surely the question we were addressing was:

Do you think that the radar distance Earth-Moon is different from the radar distance Moon-Earth?

 

[]The order is [math]\sqrt{10^{10}} \sqrt{10^{14}}=10^{12}[/math].

not really - but regardless I was kind of hoping for a helping hand to understand the reality behind that answer (whether 10e13 or 10e12)

[xyzt]

but surely the question we were addressing was:

 

not really - but regardless I was kind of hoping for a helping hand to understand the reality behind that answer (whether 10e13 or 10e12)

A probe particle will :

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

-on the return trip, it will climb out of the gravitational well of the Moon and fall into the gravitational well of the Earth. This will take a time [math]\tau_{ME}[/math]

 

The other probe particle will do the reverse trip:

-climb out of the gravitational well of the Moon and fall into the gravitational well of the Earth. This will take a time

[math]\tau_{ME}[/math]

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

 

The total time for either particle (the "radar distance") is the same, [math]\tau_{EM}+\tau_{ME}[/math], just the order of addition is reversed.

 

The key to the problem is the symmetry of the trips, resulting into both radar distances being measured to be equal . Does this make sense for you?

Edited May 24, 2013 by xyzt

[swansont]

You talk about the probe particle and its proper time, but the probe is a photon, and the proper time of a photon doesn't exist.

 

We're not talking about an inertial observer. The system is not symmetric because the two observers are in different gravitational potentials.

[xyzt]

You talk about the probe particle and its proper time, but the probe is a photon, and the proper time of a photon doesn't exist.

 

We're not talking about an inertial observer. The system is not symmetric because the two observers are in different gravitational potentials.

In GR a probe particle is considered infinitely small, so , contrary to your claim, it applies to a photon. You can see that in any book on GR.

You seem fixated on the non-inertial observers located in gravitational wells and on the fact that they are at different gravitational potentials but you fail to see that the radar beams execute round-trips against the gravitational field Earth-Moon, and these round-trips are symmetric,

Edited May 24, 2013 by xyzt

[md65536]

In GR a probe particle is considered infinitely small, so , contrary to your claim, it applies to a photon. You can see that in any book on GR.

You seem fixated on the non-inertial observers located in gravitational wells and on the fact that they are at different gravitational potentials but you fail to see that the radar beams execute round-trips against the gravitational field Earth-Moon, and these round-trips are symmetric,

That's what I was asking about in the first post, the non-inertial observers located in gravitational wells, and redshifts of one-way signals between them, and relative lengths.

 

You're now discouraging people from discussing the original topic because it apparently doesn't relate to your test probe. If the probe relates to the topic, how do you go from calculations of the probe's observations to what the observers located in gravitational wells measure?

[xyzt]

That's what I was asking about in the first post, the non-inertial observers located in gravitational wells, and redshifts of one-way signals between them, and relative lengths.

 

You're now discouraging people from discussing the original topic because it apparently doesn't relate to your test probe. If the probe relates to the topic, how do you go from calculations of the probe's observations to what the observers located in gravitational wells measure?

If you do not understand GR, you will never understand the solution. Using test probes of infinitely small mass is the standard methodology of solving problems in GR. I can recommend a few good books for you. You can continue to give me negative points out of spite, I really do not care.

[md65536]

If you do not understand GR, you will never understand the solution. Using test probes of infinitely small mass is the standard methodology of solving problems in GR. I can recommend a few good books for you. You can continue to give me negative points out of spite, I really do not care.

Sure, recommendations of good learning resources are always appreciated.

[xyzt]

Yes, we are talking GR, a subject that I am very familiar with. I can tell you in no uncertain terms that your attempts at calculating are incorrect.

Here is the deal:

A probe particle will :

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

-on the return trip, it will climb out of the gravitational well of the Moon and fall into the gravitational well of the Erath. This will take a time [math]\tau_{ME}[/math]

 

The other probe particle will do the reverse trip:

-climb out of the gravitational well of the Moon and fall into the gravitational well of the Earth. This will take a time [math]\tau_{ME}[/math]

-climb out the gravitational "well" of the Earth and fall into the gravitational "well" of the Moon. This will take a time [math]\tau_{EM}[/math]

 

The total time for either particle (the "radar distance") is the same, [math]\tau_{EM}+\tau_{ME}[/math], just the order of addition is reversed.

 

Now, for people interested in the subject, I could show the the expressions for [math]\tau_{ME}[/math] and [math]\tau_{EM}[/math]. For example:

 

[math]c \tau_{ME}=\sqrt{\frac{r_0}{r_E}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math]

 

 

 

 

 

where [math]r_E[/math] is the Earth Schwarzshild radius, [math]r_0=D[/math] and [math]r[/math] is the Earth radius. The derivation is quite complicated but I can attach a file that shows how it is done.

 

Likewise:

 

[math]c \tau_{EM}=\sqrt{\frac{r_0}{r_M}}(r_0 arctan \sqrt{\frac{r}{r_0-r}}+\sqrt{r(r_0-r)})[/math]

 

where [math]r_M[/math] is the Moon Schwarzshild radius.

 

The presence of the Schwarzschild radii encapsulates the gravitational effects on clocks.

I looked over my notes, the correct formulas are:

 

[math]c \tau_{ME}=r_0 \sqrt{1-\frac{r_E}{r_0}}=D \sqrt{1-\frac{r_E}{D}} [/math]

[math]c \tau_{EM}=r_0 \sqrt{1-\frac{r_M}{r_0}} =D \sqrt{1-\frac{r_M}{D}}[/math]

 

Sorry, I looked up the wrong section in my notes.

Edited May 24, 2013 by xyzt

[xyzt]

Sure, recommendations of good learning resources are always appreciated.

Invest in "Black Holes" by Derek Raine.

[swansont]

In GR a probe particle is considered infinitely small, so , contrary to your claim, it applies to a photon. You can see that in any book on GR.

You seem fixated on the non-inertial observers located in gravitational wells and on the fact that they are at different gravitational potentials but you fail to see that the radar beams execute round-trips against the gravitational field Earth-Moon, and these round-trips are symmetric,

 

On the contrary, I am quite aware that the photon makes a round trip. But each clock used to measure the round-trip time of the photon is not in a symmetric state. We are not trying to measure a redshift (or blueshift) of the photon. We are measuring a time with a clock.

[xyzt]

On the contrary, I am quite aware that the photon makes a round trip. But each clock used to measure the round-trip time of the photon is not in a symmetric state. We are not trying to measure a redshift (or blueshift) of the photon. We are measuring a time with a clock.

I don't think that you understand what I am telling you, the radar measurements of elapsed time are:

 

[math]\tau_{ME}+\tau_{EM}[/math] for the photon originating from the Moon radar gun

 

and

 

[math]\tau_{EM}+\tau_{ME}[/math] for the photon originating from the Earth radar gun

 

Surely you realize that the two quantities are equal (independent of any measuring technique, by virtue of the mere fact that addition is commutative). If you think otherwise, I would love to see your calculations.

Edited May 25, 2013 by xyzt

[md65536]

I don't think that you understand what I am telling you, the radar measurements of elapsed time are:

 

[math]\tau_{ME}+\tau_{EM}[/math] for the photon originating from the Moon radar gun

 

and

 

[math]\tau_{EM}+\tau_{ME}[/math] for the photon originating from the Earth radar gun

 

Surely you realize that the two quantities are equal. If you think otherwise, I would love to see your calculations.

I don't understand. For a photon with velocity c, the distance contracts to zero and the proper time is undetermined.

A clock on the moon doesn't pass through any events on Earth so it is not measuring proper times of each leg of the journey. Similar for the clock on Earth.

 

Whose clock measures Tau_EM etc and how do you figure out the round-trip times according to the Earth's clock using that?

[xyzt]

I don't understand. For a photon with velocity c, the distance contracts to zero and the proper time is undetermined.

This is false, you should definitely get the book I recommended, even better, take a class.

Here is quick proof:

 

Start with the reduced Schwarzschild solution to EFE:

 

[math]ds^2=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}[/math]

 

Light follows null geodesics [math]ds=0[/math]

 

so:

 

[math]0=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}[/math]

 

giving immediately:

 

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

 

If you integrate from [math]r_1[/math] to [math]r_2[/math] you will get the total elapsed coordinate time. I can guarantee you that it isn't zero.

 

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]

 

so:

 

 

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

 

If you integrate from [math]r_1[/math] to [math]r_2[/math] you will get the total elapsed proper time, [math]\tau_{12}=\int_{r_1}^{r_2}{\frac{dr}{c \sqrt{1-r_s/r}}}[/math]. I can guarantee you that it isn't zero, actually, it comes very close to the Newtonian value of [math]\frac{r_2-r_1}{c}[/math]. If you do the integration correctly, you will get the results that I showed in my earlier posts. It would be good if you stopped making statements that are not conform to mainstream science, especially in the context of your admitting that you do not know GR. Continuing to give me negative feedback will not make you any more correct, quite the opposite.

Edited May 25, 2013 by xyzt

[swansont]

I don't think that you understand what I am telling you, the radar measurements of elapsed time are:

 

[math]\tau_{ME}+\tau_{EM}[/math] for the photon originating from the Moon radar gun

 

and

 

[math]\tau_{EM}+\tau_{ME}[/math] for the photon originating from the Earth radar gun

 

Surely you realize that the two quantities are equal. If you think otherwise, I would love to see your calculations.

I am asking about the clock measurements of elapsed time. I have been asking about the clock measurements of elapsed time all along.

 

How does one use a photon to measure the elapsed time of its round trip?

[xyzt]

 

I am asking about the clock measurements of elapsed time. I have been asking about the clock measurements of elapsed time all along.

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

Edited May 25, 2013 by xyzt