B Demonstrating Time and Distance Effects of a Martian Signal

[Speady]

Suppose the Martian sends a signal to Earth, the final flash of which is emitted exactly four minutes later than the initial flash. In that case, with a favorable position of the planets, the final flash on Earth will also be observed neatly 4 minutes later (and 4 light minutes further than at the start of the measurement) than the initial flash. However, two satellites orbiting the Earth at 12 km / s (one in the direction of Mars and one in the opposite direction) measure a time of one hundredth of a second shorter and one hundredth of a second longer for the same signal, respectively. The question (or the test) is whether (and how) you can use a measurement (or calculation) to demonstrate or show that the distance between the start and end flash is 3000 km shorter for one satellite and 3000 km longer for the other satellite.

 

[Ibix]

The question (or the test) is whether (and how) you can use a measurement (or calculation) to demonstrate or show that the distance between the start and end flash is 3000 km shorter for one satellite and 3000 km longer for the other satellite.

Distance as measured by who?

Perhaps you should share your calculation, as you've apparently done one.

 

[PeterDonis]

Moderator's note: Thread level changed to B.

 

[PeterDonis]

with a favorable position of the planets, the final flash on Earth will also be observed neatly 4 minutes later (and 4 light minutes further than at the start of the measurement) than the initial flash.

What do you mean by "favorable position of the planets"? And what does "4 light minutes further" mean?

 

[Nugatory]

Suppose the Martian sends a signal to Earth, the final flash of which is emitted exactly four minutes later...

That's four minutes using which frame?

...In that case, with a favorable position of the planets, the final flash on Earth will also be observed neatly 4 minutes later (and 4 light minutes further than at the start of the measurement) than the initial flash.

In general, no, and I'm not sure that there is any frame in which this can be true as a special case. That's why it is so important that you properly specify the frame you are using. If I am understanding your setup properly, the Earth and Mars are not at rest relative to one another (because if it were then "4 light minutes further than at teh start of the measurement" would make no sense) so we have to worry about clock synchronization and relativity of simultaneity.

So specify the frame clearly (remembering that the speeds of the two satellites are also going to depend on the choice of frame) and you'll get better and more helpful answers.

 

[Speady]

Distance as measured by who?

Perhaps you should share your calculation, as you've apparently done one.

I can't make the calculation. If the calculation is not possible, this is a falsification of the special theory of relativity.

 

[Speady]

What do you mean by "favorable position of the planets"? And what does "4 light minutes further" mean?

When Mars, the Earth and they Sun are aligned.
The distance between the start flash and the end flash is 4 light minutes.

 

[Ibix]

I can't make the calculation. If the calculation is not possible, this is a falsification of the special theory of relativity.

No - it's your failure to specify the problem exactly. You haven't said who measures what.

The distance between the start flash and the end flash is 4 light minutes.

As measured by who?

 

[Speady]

That's four minutes using which frame?In general, no, and I'm not sure that there is any frame in which this can be true as a special case. That's why it is so important that you properly specify the frame you are using. If I am understanding your setup properly, the Earth and Mars are not at rest relative to one another (because if it were then "4 light minutes further than at teh start of the measurement" would make no sense) so we have to worry about clock synchronization and relativity of simultaneity.

So specify the frame clearly (remembering that the speeds of the two satellites are also going to depend on the choice of frame) and you'll get better and more helpful answers.

The inertial frame is that of Earth / Mars (they don't move in each other's direction).

 

[Speady]

To be clear, the Earth and the two satellites each measure a different length of time to receive the same light beam (with the same length of 72,000,000 km?). What else is the length associated with the light beams that the satellites receive?

 

[PeterDonis]

I can't make the calculation.

If you can't make a calculation, where are the numbers in your OP coming from?

If the calculation is not possible, this is a falsification of the special theory of relativity.

The fact that you can't make the calculation in no way implies that the calculation is not possible.

Claims about falsification of SR are obviously wrong, since SR has a huge amount of experimental confirmation. You need to rethink your position.

 

[Ibix]

To be clear, the Earth and the two satellites each measure a different length of time to receive the same light beam (with the same length of 72,000,000 km?)

So all you are after is the different durations of the received beams? If the Earth receives the light for four minutes, then in the Earth's frame the satellites move approximately 0.01 light seconds. Time dilation is negligible at these speeds, so the satellites receive light for four minutes plus or minus 0.01s. I don't see what your problem is.

 

[Speady]

No - it's your failure to specify the problem exactly. You haven't said who measures what.

As measured by who?

Eart and Mars measure the same duration and the same length: 4 light minutes in 4 minutes.
The satelites measure 240,01 seconds and 239.99 seconds. The question now is: wat should be the lengts measured by the satellites? (they move with a speed of 12 km/s.)

 

[PeterDonis]

When Mars, the Earth and they Sun are aligned.

They aren't at rest relative to each other in this position.

The distance between the start flash and the end flash is 4 light minutes.

Distance in what frame, for which end of the flashes (emission or reception)?

The inertial frame is that of Earth / Mars (they don't move in each other's direction).

But they are moving sideways, and at different speeds. So there is no such thing as an inertial frame in which they are both at rest.

Also, since spacetime is curved due to the Sun, a frame in which the Sun is at rest is not an inertial frame according to General Relativity. It is according to Newtonian mechanics, but you posted this in the relativity forum, so we're using relativity to analyze your scenario.

the Earth and the two satellites each measure a different length of time to receive the same light beam

If you mean the light signal coming from Mars, yes, this is true.

with the same length of 72,000,000 km

Length in what frame? As has already been shown above, there is no single inertial frame in which the Earth and Mars are both at rest.

I strongly suggest that you rework your scenario to remove all the complicating factors that you are not taking proper account of anyway. I would try a scenario with two spaceships far out in deep space, so no gravitating masses are present, exchanging light signals; and having two probes using rockets to fly circular paths around one of the spaceships.

 

[Ibix]

Eart and Mars measure the same duration and the same length: 4 light minutes in 4 minutes.
The satelites measure 240,01 seconds and 239.99 seconds. The question now is: wat should be the lengts measured by the satellites? (they move with a speed of 12 km/s.)

The lengths of the beam? 240.01 and 239.99 light seconds, obviously. This is standard Doppler calculation - you don't even need relativity at this speed.

 

[Speady]

The different durations are clear and no problem. The problem is how to determine the corresponding lengths. The question is: what does the calculation look like that changes the light beam of 72,000,000 km into a light beam of 72,003,000 km and one of 71,997,000 km.

 

[Ibix]

The different durations are clear and no problem. The problem is how to determine the corresponding lengths. The question is: what does the calculation look like that changes the light beam of 72,000,000 km into a light beam of 72,003,000 km and one of 71,007,000 km.

The source is moving differently in the different frames. It was at some distance $d$ when it started emitting and at $d\pm0.01\mathrm{light\, seconds}$ when it finished.

Edit: You mean 71,997,000 km, I presume

 

[Speady]

This does not apply to Earth and Mars. There is a relative speed between Mars and the satellites: 12 km / s.

 

[Nugatory]

The inertial frame is that of Earth / Mars (they don't move in each other's direction).

Move “in each other’s direction” or “move relative to one another”? I‘m assuming that you meant the latter, which makes the problem much easier to analyze. And since then I’ve seen your clarification that when you said

the final flash on Earth will also be observed neatly 4 minutes later (and 4 light minutes further than at the start of the measurement) than the initial flash.

the bit in parentheses was intended to say that the length of the pulse was four light-minutes.

The last ambiguity is whether the “distance between the start and the end flash” for each satellite is the distance using the frame in which that particular satellite is at rest, or the distance using the frame in which Mars and the Earth are at rest; presumably you mean the former because it’s trivially 4 light-minutes using the latter.

OK, with all that specification out of the way so we agree about the problem...

This is actually pretty easy. Using the frame in which the Earth and Mars are at rest, write down the t and x coordinates of six events:
1) the leading edge of the pulse reaches the first satellite
2) the trailing edge of the pulse reaches the first satellite
3) the leading edge of the pulse reaches the second satellite
4) the trailing edge of the pulse reaches the second satellite
5) the leading edge of the pulse leaves Mars
6) the trailing edge of the pulse leaves Mars
Next use the Lorentz transformations to calculate the coordinates of events 1,2,5,6 using a frame in which the first satellite is at rest; and again to calculate the coordinates of events 3,4,5,6 using a frame in which the second satellite is at rest.

Now you can read off from these coordinate values the distance traveled by both edges of the pulse, the width of the pulse, the time between the arrival of the edges of the pulse, as measured by each satellite. I have no idea if it will turn out to be the numbers you’re suggesting - 3000 km shorter (shorter than what? Than as measured using the frame in which Earth and Mars are at rest?) for one satellite, 3000 km longer for the other - but whatever it is, that’s how we calculate it.

 

[Ibix]

This does not apply to Earth and Mars. There is a relative speed between Mars and the satellites: 12 km / s.

I did miss the third frame from my response - the Earth's rest frame. In that one, the beam length is four light minutes because in that frame the source is at rest and it emits for four minutes.

The detailed calculation is in Nugatory's post.

 

[Speady]

Now you can read off from these coordinate values the distance traveled by both edges of the pulse, the width of the pulse, the time between the arrival of the edges of the pulse, as measured by each satellite. I have no idea if it will turn out to be the numbers you’re suggesting - 3000 km shorter (shorter than what? Than as measured using the frame in which Earth and Mars are at rest?) for one satellite, 3000 km longer for the other - but whatever it is, that’s how we calculate it.

There should be a calculation in the SR that extends a length of 72,000,000 km 3000 km at a speed of 12 km / s. It does not exist.

 

[Ibix]

There should be a calculation in the SR that extends a length of 72,000,000 km 3000 km at a speed of 12 km / s. It does not exist.

Yes it does - Nugatory laid it out above.

There is a distinction between you not understanding something and it not existing.

 

[Speady]

I have no idea if it will turn out to be the numbers you’re suggesting - 3000 km shorter (shorter than what? Than as measured using the frame in which Earth and Mars are at rest?) for one satellite, 3000 km longer for the other - but whatever it is, that’s how we calculate it.

I have: 72,000,000 km will remain 72,000,000 km. De speed will be 300,988 km/s and 300,012 km/s.

 

[PeterDonis]

The different durations are clear and no problem. The problem is how to determine the corresponding lengths.

Lengths of what?

The satellites (and Earth) are not measuring the lengths of anything. They are measuring the time it takes for the entire signal to arrive.

In some frames the satellites are moving, so they cover distance while they are receiving the signal; but that distance isn't the length of anything associated with the signal itself.

I think you need to think more carefully about what you mean by "length".

 

[PeterDonis]

I have: 72,000,000 km will remain 72,000,000 km. De speed will be 300,988 km/s and 300,012 km/s.

You said you weren't capable of doing a calculation. But you're giving numbers here, and you've given numbers before in this thread. Where are those numbers coming from?

 

[Speady]

Lengths of what?

The satellites (and Earth) are not measuring the lengths of anything. They are measuring the time it takes for the entire signal to arrive.

In some frames the satellites are moving, so they cover distance while they are receiving the signal; but that distance isn't the length of anything associated with the signal itself.

I think you need to think more carefully about what you mean by "length".

I think the length is not changing. I think te speed perceaved by the satellites is 300,998 km/s and 300,012 km/s.

 

[PeterDonis]

you can read off from these coordinate values the distance traveled by both edges of the pulse, the width of the pulse, the time between the arrival of the edges of the pulse, as measured by each satellite.

You can read off times from the coordinates in the appropriate frames, but that's just because the time coordinate values in those frames just reflect the proper time of the appropriate object (Mars or one of the satellites).

You can't, however, read off distances from these coordinate values, because none of them are for pairs of events that either (a) are simultaneous in any of the given frames, or (b) take place at two objects that are at rest relative to each other.

The root problem is that the OP has not thought through what "distance" and "length" means, and the limitations of those concepts in the scenario he is posing.

 

[PeterDonis]

I think the length is not changing. I think te speed perceaved by the satellites is 300,998 km/s and 300,012 km/s.

Enough is enough. You are claiming that SR is wrong, but you cannot or will not give any actual math to back up your claims.

Thread closed.

 

[Nugatory]

I think the length is not changing. I think te speed perceaved by the satellites is 300,998 km/s and 300,012 km/s.

The speed of light as perceived by each satellite (more precisely, using a frame in which the satellite is at rest) is $c$. The speed of light is $c$ in all inertial frames.