Straight Dope Message Board > Main My Problems With Relativity
 Register FAQ Calendar Mark Forums Read

#551
05-15-2012, 11:50 AM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam It will take a full 1,000,000,000 ns (1 second) to transmit the 1,000,000,000 cycles. At 1,000,003,094 Hz, those cycles will be reflected in only 999,996,906 ns, as you apparently agree, given the bolded part of your quote. Since the source is only 10 ns away, you have those 1,000,000,000 cycles finished being reflected before they have all been transmitted. You're violating causality. How can you expect anyone to take you seriously when you write stuff like this? You continue to be clearly, completely, and unequivocally wrong.
If you can't come up with anything better than that, you have already accepted that you have lost the argument.
The travel time to the target is irrelevant. The one second pulse hits the target. After reflection the pulse will be less than one second.

Goodbye.
#552
05-15-2012, 02:11 PM
 John Cazale Guest Join Date: May 2012
Tom, I encourage you to take another look at Bunny's post a few pages ago:

Quote:
 Originally Posted by BunnyTVS So far you have shown flaws in your understanding of: Momentum Elasticity Velocity Acceleration Special Relativity Your previous arguments.
Whenever someone points out an obvious flaw in your understanding of introductory physics, you either 1)dismiss the error as a minor technical slip-up of yours or 2)just ignore them. Off the top of my head, the three most obvious errors repeatedly pointed out to you but which you have entirely ignored were 1)

Quote:
 If the puck was a beach ball, which is far easier to compress, of course its velocity would be higher than 10 Kph, if it was totally compressed and then reformed, its speed could be 20 Kph.
This is unambiguously exactly the opposite of very basic physics (and intuition to boot).

error 2)

Quote:
 the Wiki cart example is wrong - as proved by turning it on its side. If it was correct, we would have perpetual motion. [added link]
This betrays all sorts of misunderstanding. Again, as others have pointed out, what you mean by "turning it on its side" is to introduce another force, namely, the earth's gravitational force. Of course the model would play out differently if we add in new forces! This observation does not undermine the accuracy of the diagram whatsoever.

error 3), in my mind the most basic

Quote:
 London has a relative velocity of zero relative to Bath or any other point in the world you care to mention. There is your proof. If I travel from London to Bath, I can measure the acceleration of Bath can I? Or can I measure the accleration of Bath from London?
Again, even intuitively, this is clearly wrong. Just think about it - how can the velocity at the equator and at the pole be the same?

I am very much a layman, so I apologize for any errors I may have made myself. My point is that even someone with a very rudimentary knowledge of basic physics can see the gaping holes in your reasoning. Your unwillingness to address these errors (among many others) proves either that you 1) aren't arguing in good faith, or 2) are suffering from a serious case of the Dunning-Kruger effect, as has already been mentioned, or perhaps some combination of the two.

I'll be surprised and impressed if you actually address the errors in the quotations I posted.

Last edited by John Cazale; 05-15-2012 at 02:16 PM.
#553
05-15-2012, 07:05 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 6,893
Quote:
 Originally Posted by tomh4040 If you can't come up with anything better than that, you have already accepted that you have lost the argument. The travel time to the target is irrelevant. The one second pulse hits the target. After reflection the pulse will be less than one second.
Painted yourself into a corner, I see.
#554
05-16-2012, 04:30 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by naita In the RFR of Earth the speed is different, yes, as I showed in my calculations. As viewed from any IFR, they are not.
We agree on this point.

Quote:
 No, it has changed its velocity. The speed is the same. If you examine the situation properly, you'll also find that the radial velocity is the same in the moment of emision and the moment of reflection.
I was talking about a linearly accelerated frame in that passage, with the light/radar pulse being in the same direction as the line of travel.

Quote:
 No, the speed changes, and the wavelenghts change correspondently, leaving the frequency the same.
Read the explanation and the maths again. If the radar increases its velocity, then the same number of wavecrests are now in a shorter duration pulse. Velocity and frequency have changed, so the wavelength stays the same. This can be seen in any wave phenomena.

Quote:
 For that to make sense your reflector would emit twice the pulses it received. The only way frequency changes is if the distance between emitter and receiver changes Your 10 Hz transmitter is sending out 10 pulses every second, the reflector is sending out 20 pulses every second. The distance between them is constant.
There are two ways the frequency can change. You have mentioned one. The other one is if the velocity changes.
The reflector does not reflect 20 cycles in a second. If you remember, I fell into that trap with ZenBeam. The reflector reflects 10 cycles, yes. But it reflects them in half a second not one second, so the frequency is doubled.

Quote:
 The number of pulses transmitted is 10*t where t is elapsed time. The number of pulses reflected is 20*(t-tt) where tt is travel time between transmitter and reflector. (Valid from t=tt and onwards.) Say the travel time is 10 seconds. After 10 seconds the transmitter has sent out 100 pulses, and none have yet been reflected. After 20 seconds the transmitter has sent out 200 pulses, and 200 pulses have been reflected, which is odd, since 100 pulses are en route between transmitter and reflector. After 40 seconds the transmitter has sent out 400 pulses, and 600 pulses have been reflected, which is frankly mindboggling.
This is almost word for word what ZenBeam put, and the answer is the same. Paragraph two - correct.
Paragraph three - incorrect. 200 pulses have not been reflected, only 100 have been reflected, but they have been reflected in half the time.
Paragraph four - same as three. You are confusing pulses (cycles - period) with cycles per second, which is Hertz.

Quote:
 No, it doesn't, it results in a change in wavelength. That's what happens in all other situations where the speed of a wave changes.
In all other situations; sound or waves on water for example, the frequency changes. Listen to an outdoor orchestra in a varying wind, throw pebbles in a pond and a flowing river. You will hear/see for yourself.
Quote:
 Frequency change requires the distance between source and receiver to change.
That is one of the mechanisms for a change of frequency, the other is a change of speed.

Quote:
 If the simple calculations based on your 10 Hz example don't convince you you're wrong I'll draw and explain this visually later.
Please do it now. You, and all the other posters have seen my maths. Let us all see yours.

Quote:
 Originally Posted by John Cazale Tom, I encourage you to take another look at Bunny's post a few pages ago: Whenever someone points out an obvious flaw in your understanding of introductory physics, you either 1)dismiss the error as a minor technical slip-up of yours or 2)just ignore them. Off the top of my head, the three most obvious errors repeatedly pointed out to you but which you have entirely ignored were 1) This is unambiguously exactly the opposite of very basic physics (and intuition to boot). error 2) This betrays all sorts of misunderstanding. Again, as others have pointed out, what you mean by "turning it on its side" is to introduce another force, namely, the earth's gravitational force. Of course the model would play out differently if we add in new forces! This observation does not undermine the accuracy of the diagram whatsoever. error 3), in my mind the most basic Again, even intuitively, this is clearly wrong. Just think about it - how can the velocity at the equator and at the pole be the same? I am very much a layman, so I apologize for any errors I may have made myself. My point is that even someone with a very rudimentary knowledge of basic physics can see the gaping holes in your reasoning. Your unwillingness to address these errors (among many others) proves either that you 1) aren't arguing in good faith, or 2) are suffering from a serious case of the Dunning-Kruger effect, as has already been mentioned, or perhaps some combination of the two. I'll be surprised and impressed if you actually address the errors in the quotations I posted.
Of course I will address the errors (?), but in return, you will have to answer my questions about the apparent incompatibility of the Sagnac effect with radar in my last few postings.
Point 1. The beach ball. I was using poetic licence. A beach ball has a very low weight to size ratio, the characteristics are unknown, and air resistance plays a big part. Substitute for a ball of known characteristics, and use that. The characteristics of this ball are that is heavy, so air resistance can be (nearly) ignored, and it reforms at 10kph. When this ball is hit by a bat and therefore compressed, it will compress to half its size in the direction it is hit, and then reform again. The bat moving at 10kph (WRT the ground) hits the ball, which compresses while still in contact with the bat. The bat is still moving at 10kph and the ball is fully compressed. Now bat and ball are both moving at 10kph WRT the ground. The ball reforms to its original shape while the surface of the ball is still in contact with the bat and is attempting to move towards the bat at 10kph (WRT the bat). As the two surfaces are in contact, this means that the centre of mass of the ball is pushed away from the bat at 10kph. The bat is still moving at 10kph WRT the ground, so the ball is now moving at 20kph WRT the ground.

We have moved on since then. Now you tell me why I am wrong in saying the Sagnac effect is incompatible with radar. See my posting #547.
#555
05-16-2012, 06:28 PM
 naita Guest Join Date: Jun 2002
Let's again place the transmitter at A and the reflector at B.

Let's nail down our description of the rotating reference frame co-rotating with A. In this reference frame neither A or B moves, and there's a constant distance between them. The speed of light differs in the AB and BA direction due to it being a rotating reference frame.

Let's look at just two wavefronts, one cycle, emitted from A at 25GHz, i.e. 40 ps apart.

If we start transmitting at t=0, the second wavefront will be emitted at t=40 ps.

We can call travel time from A to B tAB.

The first wavefront will hit B at tAB, the second wavefront will hit B at tAB + 40 ps

As they hit they are reflected. The first wavefront is reflected at tAB, the second wavefront is reflected at tAB + 40 ps

Travel time from B to A is different from the travel time from A to B. We can call travel time from B to A tBA.

The second wavefront will return to B at tAB + tBA + 40 ps

They're received at A on their return 40 ps apart, which is 25GHz.

Now I suspect you'll disagree that A and B are not moving, but in the reference frame co-rotating with A they are still. That's what's meant by a co-rotating reference frame.

One can pick an infinity of other reference frames, both rotating, accelerated and inertial, where A and B do move, but the only one that makes sense to examine (the other's will give equivalent results, just with horrible math) is an inertial one co-moving with A and B's center of rotation. In that frame A and B do have acceleration and velocity, but as in all other inertial frames of reference, c is constant and the radial velocity between co-rotating points A and B is 0.

Now the following is just an animation I made for my own amusement which shows "photons" moving with or against the direction of rotation, it's not meant to really enlighten anyone in this thread. But it's as good a place to share it as any: http://www.geogebratube.org/student/m9696 (requires up to date Java, and in my case, use of my second choice browser)
#556
05-16-2012, 07:27 PM
 John Cazale Guest Join Date: May 2012
I'm sorry, Tom, but you are clearly being disingenuous. You only "explained" one of your obvious errors. Is this

Quote:
 London has a relative velocity of zero relative to Bath or any other point in the world you care to mention.
poetic license? Give me a break.

Again, the lack of knowledge exhibited in that quotation means you just don't have the ability (at this time) to make sense of something like relativity.

Lastly, a few times throughout the discussion you've expressed interest in converting others to your view. If you really want the ether theory to gain traction I'd encourage you yourself to stop talking about it, as you are supremely unconvincing.

Last edited by John Cazale; 05-16-2012 at 07:27 PM.
#557
05-17-2012, 03:24 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by John Cazale I'm sorry, Tom, but you are clearly being disingenuous. You only "explained" one of your obvious errors. Is this poetic license? Give me a break. Again, the lack of knowledge exhibited in that quotation means you just don't have the ability (at this time) to make sense of something like relativity. Lastly, a few times throughout the discussion you've expressed interest in converting others to your view. If you really want the ether theory to gain traction I'd encourage you yourself to stop talking about it, as you are supremely unconvincing.
I answered only one point, and I notice you did not disagree with the explanation. Now I want you to answer mine.
#558
05-17-2012, 04:25 PM
 John Cazale Guest Join Date: May 2012
Quote:
 Originally Posted by tomh4040 I answered only one point, and I notice you did not disagree with the explanation. Now I want you to answer mine.
This is ridiculous. I've already admitted I'm a layman, and certainly you can admit that your question isn't basic physics in the same way that the definition of velocity is. So you might say that our positions in this matter lack symmetry.

Again: I know what velocity is. You seem to not. If you don't understand velocity, you couldn't possibly understand the physics you're asking about.

Last edited by John Cazale; 05-17-2012 at 04:29 PM.
#559
05-17-2012, 05:50 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by tomh4040 . . . If the radar increases its velocity . . .
Isn't this the core of the issue? According to SR, this doesn't happen. It's also never been observed to happen.
#560
05-20-2012, 12:03 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by naita Let's again place the transmitter at A and the reflector at B. Let's nail down our description of the rotating reference frame co-rotating with A. In this reference frame neither A or B moves, and there's a constant distance between them. The speed of light differs in the AB and BA direction due to it being a rotating reference frame. Let's look at just two wavefronts, one cycle, emitted from A at 25GHz, i.e. 40 ps apart. If we start transmitting at t=0, the second wavefront will be emitted at t=40 ps. We can call travel time from A to B tAB. The first wavefront will hit B at tAB, the second wavefront will hit B at tAB + 40 ps As they hit they are reflected. The first wavefront is reflected at tAB, the second wavefront is reflected at tAB + 40 ps Travel time from B to A is different from the travel time from A to B. We can call travel time from B to A tBA. The first wavefront will return to A at tAB + tBA. The second wavefront will return to B at tAB + tBA + 40 ps They're received at A on their return 40 ps apart, which is 25GHz.

Your posting contains another mistake, but reading it made me realize that there is a mistake in my work also. I will get to that later, but now to yours.
Not a quote, but you said “On the outward journey, to the reflector, the radar pulse is traveling at say 300,000,000 M/s, with a time between wavecrests of 40 ps. On the return journey, at say 200,000,000 M/s, the wavecrests are still 40 ps apart.” That is what you put ( I may have the velocity/direction wrong, but the maths will still be correct). You also said that the frequency is 25 Gig in both directions.

For the outward pulse, the wavecrests are 40 ps apart in time, so they have moved 0.003 M in that time.
This corresponds to a frequency of : F = V / W = 3e8 / 0.012 = 25 Gig

They are reflected also at 40 ps apart. Now the velocity is not 3e8 but is 2e8 (the return pulse). The wavecrests have moved 0.002 M in that time.
This corresponds to a frequency of : F = V / W = 2e8 / 0.012 = 16,666,666,666.66 Gig

This took me by surprise, and I looked harder at my previous postings. There is a mistake there, but as I am about to leave for a well earned holiday in Crete, you will have to wait until my return for my announcement. Of course you are all allowed to speculate in my absence.

A quick answer to Trinopus. Relativity says that the speed of light is a constant (c) in any and all IFRs. To be constant in an IFR, it cannot be constant in an RFR. It is for that very reason that a ring laser (Sagnac interferometer) works. Read up on that.

To John Cazale. You consider yourself a layman, but that does not stop you disagreeing with the points I have made in this discussion. You cannot hide behind a layman's tag when it suits you not to answer a question, and yet come out from behind it to disagree with me. Read back through these posts, you will find that I do know the difference between speed and velocity. You may be making that remark based on the observation that I sometimes use the word speed, and at other times use the word velocity. Sometimes they are interchangeable, sometimes they are not.
#561
05-20-2012, 02:26 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 6,893
Quote:
 Originally Posted by tomh4040 They are reflected also at 40 ps apart. Now the velocity is not 3e8 but is 2e8 (the return pulse). The wavecrests have moved 0.002 M in that time. This corresponds to a frequency of : F = V / W = 2e8 / 0.012 = 16,666,666,666.66 Gig This took me by surprise, [...]
Wow. Just ... amazing. You don't even understand the relationship between period and frequency. This is so basic...

From Wikipedia:
Quote:
 Frequency is the number of occurrences of a repeating event per unit time. It is also referred to as temporal frequency. The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency. [...] The period, usually denoted by T, is the length of time taken by one cycle, and is the reciprocal of the frequency f: T = 1 / f
If the period is 40 ps, the frequency is the reciprocal of that, 1/(40 ps), or 25 GHz.

Here's another, simpler page that be at more your level:
Quote:
 Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity. Period is a time quantity. Frequency is the cycles/second. Period is the seconds/cycle. [...] Mathematically, the period is the reciprocal of the frequency and vice versa. In equation form, this is expressed as follows. period = 1 / frequency frequency = 1 / period

Last edited by ZenBeam; 05-20-2012 at 02:27 PM.
#562
05-20-2012, 04:56 PM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,012
I have just got to say that the difference between RFRs and IFRs has never been so well explained before. I have two years of college physics under my belt, but it was a while ago. Thanks to naita for pointing out that radial velocity is what Doppler shifting is all about, even in RFRs. I made a remark about a month ago regarding my enjoyment of the thread and I have since kept pace reading along. Let's see if I have my facts straight:
• There is no Doppler shift observed in a ring laser in a RFR.
• There is a fringe pattern observed in a RFR.
• The RFR observer sees a difference in c.
• The IFR observer sees no difference in c.

I am on track here?
#563
05-21-2012, 02:28 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by MonkeyMensch I have just got to say that the difference between RFRs and IFRs has never been so well explained before. I have two years of college physics under my belt, but it was a while ago. Thanks to naita for pointing out that radial velocity is what Doppler shifting is all about, even in RFRs. I made a remark about a month ago regarding my enjoyment of the thread and I have since kept pace reading along. Let's see if I have my facts straight: There is no Doppler shift observed in a ring laser in a RFR. There is a fringe pattern observed in a RFR. The RFR observer sees a difference in c. The IFR observer sees no difference in c. I am on track here?

Yes. As viewed from any IFR the fringe pattern is the result of a difference in path length. As viewd from the RFR the fringe pattern is the result of a difference in the speed of light.
#564
05-21-2012, 02:40 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by tomh4040 For the outward pulse, the wavecrests are 40 ps apart in time, so they have moved 0.003 M in that time. This corresponds to a frequency of : F = V / W = 3e8 / 0.012 = 25 Gig They are reflected also at 40 ps apart. Now the velocity is not 3e8 but is 2e8 (the return pulse). The wavecrests have moved 0.002 M in that time. This corresponds to a frequency of : F = V / W = 2e8 / 0.012 = 16,666,666,666.66 Gig
The usual symbol for frequency is f, the symbol for speed/velocity is v, the symbol for wavelength is usually the greek letter lambda, but if you have to use lating letters, W is a poor choice. This does not matter to the math of course, but it makes things easier to read.

Moving at 3e8 m/s 40 ps apart we get a wavelength of 0.012 m as you correctly put in your formula. I don't know where you get 0.003 M (and the symbol for meters is m, not M.), unless it's how far the waves move in 100 ps.

Moving at 2e8, we get l = v*t = 2e8*40e-12 = 0.008 m. Again it looks like you've calculated how far they travel in 100ps, but you've kept the wavelength from the first situation, which is wrong. If the first wavecrest is reflected at t = 0, the second is reflected at t = 40 ps, the first wavecrest is then gets a head start of 0.008 m, which is the wavelength.

Using this correct wavelength in f= v/l, we get f = 2e8/0.008 = 25 GHz .
#565
05-21-2012, 03:12 PM
 Maeganspop Guest Join Date: May 2012
TV Signal

Hi, I'm new to this board and this thread caught my attention.

I haven't read every page and this may have already been talked about, what if the space ship traveling to a distant star was receiving a television signal from Earth.

The space ship is on a launch pad and has a TV set with a signal from mission control. The astronauts are seeing on the TV everything in real time and it's normal.

The space ship blasts off and starts accelerating away from Earth. As they accelerate does the action on the TV slow down? Since the signal is traveling at the speed of light and they are accelerating to a speed that is only 98% of the speed of light, then they will never pass the signal.

I know that it would take a very powerful signal to reach the space ship but, let's say it is powerful enough. Would the action on the TV slow down the faster the ship goes? When the ship starts slowing down, would the action on the TV start speeding back up?

When the ship returns to Earth would the action on the TV start going really fast the closer they get to Earth?

Like I said, I'm sorry if this has already been discussed. It just popped into my brain and I just had to ask.
#566
05-21-2012, 03:34 PM
 naita Guest Join Date: Jun 2002
You can simplify this by considering the case where the travellers are just looking at the earth through an incredibly powerful telescope.
Just by moving away from the Earth you are getting a steadily increasing delay in your visual impression of happenings on Earth, in effect a slowing down of the observed action. And yes, the faster you move the slower the action. Slowing down will speed it up until it's back to normal when you're at rest relative to the Earth.

Moving towards the Earth the action will speed up, but the speed of the action will depend on the speed of the spaceship relative to Earth, not the distance.

At very high speed the colour will be off, while if you stick to the TV-example you need to keep adjusting the frequency.
#567
05-21-2012, 03:41 PM
 AndrewL Guest Join Date: Sep 2001
It'll be a bit more complex than just the action slowing down.

First, the carrier frequency that the TV signal is being transmitted on will shift. As the ship accelerates, the signal coming from Earth will appear to the people on the ship to be blue-shifted, with the signal apparently decreasing in frequency. On the other hand, as they approach the speed of light, time dilation will cause the people on the ship to experience time more slowly than those on the ground, which will make it seem like the signal is increasing in frequency to the people on the ship. I'm not sure which of these effects will dominate over which range of speed.

In either case, the ship's receiver will need to adjust what frequency it is listening for the signal on, as they apparent frequency in the reference frame of the ship shifts.

Secondly, the amount of time it takes for one frame of the TV signal to be transmitted will change. If the TV signal is being transmitted as a plain old analog TV signal, the TV on the ship will very quickly be unable to interpret the signal at all - analog TV signals require fixed timing of various parts of the signal in relation to each other, if time is passing at different rates at the transmitter and receiver the signal will be unintelligible.

If we assume the receiver is capable of handling the frequency shift, and is smart enough to deal with the timing shifts in the transmitted signal, then yes, the action will seem to slow down or speed up, as it will take more or less time to transmit each frame of video.
#568
05-21-2012, 03:49 PM
 Maeganspop Guest Join Date: May 2012
Thanks for the replies. I thought that the action would slow down the faster that the ship traveled away from Earth.

Both replies I read just confirmed it for me. I didn't say it in my post, but I was thinking that the TV would have to automatically adjust since the frequencies would change the faster the ship went.

Thanks again.
#569
05-21-2012, 05:15 PM
 The Hamster King Charter Member Join Date: Jun 2000 Location: Los Angeles Posts: 8,754
Quote:
 Originally Posted by AndrewL On the other hand, as they approach the speed of light, time dilation will cause the people on the ship to experience time more slowly than those on the ground, which will make it seem like the signal is increasing in frequency to the people on the ship.
Actually, no. From the perspective of Earth, time is flowing more slowly on the ship. But from the perspective of the ship, time is flowing more slowly on Earth. As the ship approaches light speed, the frequency of the signal from Earth will continue to drop (assuming that the ship's thrust is kept low enough to minimize the time dilation from the acceleration itself).
#570
05-21-2012, 08:15 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
By the "clock paradox," if I fly out into space, very fast, and then turn around and return, a large amount of time may have elapsed on earth. I might only age 20 years, but come back to earth and find that they're celebrating the year 2550 (the twenty-fourth-and-a-half century!)

If my receiver was any good, I'd have centuries of old tv shows archived on my ship's computer.

And I thought it was hard keeping up with my soaps as it was!
#571
05-21-2012, 10:35 PM
 John Cazale Guest Join Date: May 2012
Quote:
 I do know the difference between speed and velocity. You may be making that remark based on the observation that I sometimes use the word speed, and at other times use the word velocity. Sometimes they are interchangeable, sometimes they are not.
More evasion, Tom. Again, you said:

Quote:
 London has a relative velocity of zero relative to Bath or any other point in the world you care to mention.
You can use "speed" or "velocity" - either way, that sentence is obviously dead wrong, and you have no excuse. A layman like myself can challenge your understanding because it's clear you have less than a layman's grasp of the material.
#572
06-10-2012, 12:03 PM
 tomh4040 Guest Join Date: Oct 2011
I should have included this in my last posting. I agree with you on your point that A and B are not moving WRT each other, and I have always held that position. This is from an earlier posting :- “The radar pulse is c relative to an IFR, not to the Earth. According to relativity the target is moving in a non linear fashion relative to that IFR, and is therefore moving away from or towards the radar pulse.”

I actually made two mistakes. One which I made in that reply and did not spot until after I had posted it was that I had used the outbound wavelength for calculating the reflected frequency. The mistake which I knew I had made after reading your posting was in using the conveyor belt analogy. If a conveyor belt running at 1 M/s is loaded with packages every 100 milliseconds, they will be on the belt at 0.1M intervals. This corresponds to :-
F = V / W = 1 / 0.1 = 10
If the belt is speeded up to 2 M/s, the packages will still be at 0.1M intervals, but will now pass a particular point every 50 ms :-
F = V / W = 2 / 0.1 V = 20 : The frequency has doubled., while the wavelength has stayed the same.

However, this is not the correct way to look at light. This is a more correct analogy. That conveyor belt running at 1 M/s is feeding another belt which is running at 2 M/s. The packages drop off belt 1 and onto belt 2 (this corresponds to the radar being reflected and changing speed). As the packages transfer onto belt 2, they still drop onto it at 100ms intervals, but as the belt has moved 0.2 M in that time, they are now spaced apart 0.2 M.
This corresponds to :-
F = V / W = 2 / 0.2 = 10 . The frequency has not changed, the wavelength has changed.
I got that wrong, and I do admit my mistakes.
Well done to you, naita. You just worked at it, using maths and logic, and did not resort to a smear campaign or personal attacks as some (most) people on this forum have done.

The upshot of this is that although the frequency does not change with the change of speed when light/radar changes direction from E to W or vice-versa, the wavelength does change. I will look into that to see if it throws up any anomalies.

I was hoping to be able to bring this to a conclusion without needing to set up a new experiment to prove it. That goal is eluding me. One way to prove whether relativity is correct or not, is to make into a reality the Wiki setup of splitting a light beam and sending it around the globe. This does not have to be done at the equator, any line of latitude will do. It can be done near the north or south pole, with a length of fibre optic cable running round it to form a ring laser. Lay the fibre optic cable in a circle of 10 Km radius centred on the pole. This marks out a line of latitude which is 62,840 M long and rotates in 24 hours, so the speed is around 5e-5 radians per second (or 0.7273 M/s radial velocity) which is well within the operational capability of a ring laser. Split a pulse of light and send it round the loop in opposite directions. If the pulses appear back at the entry/exit point together, relativity is wrong.
#573
06-10-2012, 12:16 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by John Cazale I'm sorry, Tom, but you are clearly being disingenuous. You only "explained" one of your obvious errors. Is this poetic license? Give me a break.
Here is your big break. Take a tape measure or any other measuring device, and measure the distance form London to Bath. Wait a day or a year or however long you like, and measure again. Barring tectonic plate movements, the measured distance will not have changed. It will be the same whenever you measure it. Report back to this forum what change in distance you have found, and what the relative speed/velocity is between London and Bath.
#574
06-10-2012, 03:28 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
In physics, "velocity" is a vector. It has a magnitude, and a direction.

Since London and Bath are moving around the earth's axis, the direction of their motion is constantly changing. (Yes, it is always to the "east," but....) Their velocities are changing with respect to each other, also, for the same reason.

An object in a circular orbit never gets any closer or farther away from its primary, but it is still always *moving* with respect to its primary.
#575
06-10-2012, 05:38 PM
 Grey Guest Join Date: Sep 2002
Quote:
 Originally Posted by tomh4040 Here is your big break. Take a tape measure or any other measuring device, and measure the distance form London to Bath. Wait a day or a year or however long you like, and measure again. Barring tectonic plate movements, the measured distance will not have changed. It will be the same whenever you measure it. Report back to this forum what change in distance you have found, and what the relative speed/velocity is between London and Bath.
Spectacular! N ow, where's the second observer?
#576
06-10-2012, 08:53 PM
 Andy L Member Join Date: Oct 2000 Posts: 1,992
Quote:
 Originally Posted by Trinopus In physics, "velocity" is a vector. It has a magnitude, and a direction. Since London and Bath are moving around the earth's axis, the direction of their motion is constantly changing. (Yes, it is always to the "east," but....) Their velocities are changing with respect to each other, also, for the same reason. An object in a circular orbit never gets any closer or farther away from its primary, but it is still always *moving* with respect to its primary.
Yep. An even easier situation to think about is a circular racetrack - the cars never get any closer or farther from the central point, but they are certainly moving relative to the center. If you consider the center point to be at (0,0) and the race track to be of radius r, then for a car moving at a constant speed of v, the car's x coordinate is r*cos(v/r*t) and the car's y coordinate is r*sin(v/r*t) - so radial distance never changes, and thus the radial velocity is zero - but the total velocity is never zero.
#577
06-11-2012, 04:06 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by tomh4040 The upshot of this is that although the frequency does not change with the change of speed when light/radar changes direction from E to W or vice-versa, the wavelength does change. I will look into that to see if it throws up any anomalies.
Spoiler! It doesn't.

Quote:
 I was hoping to be able to bring this to a conclusion without needing to set up a new experiment to prove it. That goal is eluding me. One way to prove whether relativity is correct or not, is to make into a reality the Wiki setup of splitting a light beam and sending it around the globe. This does not have to be done at the equator, any line of latitude will do. It can be done near the north or south pole, with a length of fibre optic cable running round it to form a ring laser. Lay the fibre optic cable in a circle of 10 Km radius centred on the pole. This marks out a line of latitude which is 62,840 M long and rotates in 24 hours, so the speed is around 5e-5 radians per second (or 0.7273 M/s radial velocity) which is well within the operational capability of a ring laser. Split a pulse of light and send it round the loop in opposite directions. If the pulses appear back at the entry/exit point together, relativity is wrong.
You're over-complicating things. Any of the existing interferometers sensitive enough to measure the Earth's rotation proves the hypothesis in question wrong. Light speed is not constant relative to the surface of the Earth. The results are however consistent with the relativity-compatible hypothesis of light speed being constant relative to an inertial observer.
#578
06-13-2012, 02:42 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by naita Spoiler! It doesn't.
Actually it does, but the difference between the two wavelengths is very difficult to measure. I am trying to find a way to amplify that difference.

Quote:
 Originally Posted by naita You're over-complicating things. Any of the existing interferometers sensitive enough to measure the Earth's rotation proves the hypothesis in question wrong.
Please point me to a ring laser/interferometer result which proves that light speed is not constant WRT the Earth.

Quote:
 Originally Posted by naita Light speed is not constant relative to the surface of the Earth. The results are however consistent with the relativity-compatible hypothesis of light speed being constant relative to an inertial observer.
That is not a relativity compatible hypothesis, it is a central plank of relativity.
Using the setup I described (setting up a ring laser round the pole), you would expect the ring laser to show rotation then.
#579
06-13-2012, 05:34 PM
 Senegoid Guest Join Date: Sep 2011
I'm no physicist, but I drop in on this thread once every few months or so just to see where it's going.

May I threadshit? I'll put it in a spoiler for those who don't want to see it.

SPOILER:

This thread has been going nowhere as far as I can tell, that's where. How long has it been since anything actually new has come up here?

tomh4040 and others (ZenBeam, Trinopus, et many al) have simply been orbiting each other. If I measured the distance from tomh4040's arguments to any of the others today, and measured again a day, week, or year from now, will there be any change in their relative distances from one another? Do any of the arguments and rebuttals, for the last umpty posts, have any relative velocity to one another?

I can't help but wonder why y'all don't just let it rest. It's not critically urgent for everybody in the world to understand relativity. I don't very much. London and Bath will still keep orbiting around their axis no matter what tomh4040's time-invariant beliefs are. Why is everybody investing all this effort to fix that?

Just wond'rin.

Last edited by Senegoid; 06-13-2012 at 05:36 PM. Reason: minor wording clarifications
#580
06-13-2012, 06:43 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 6,893
Quote:
 Originally Posted by Senegoid I'm no physicist, but I drop in on this thread once every few months or so just to see where it's going. May I threadshit? I'll put it in a spoiler for those who don't want to see it. SPOILER: This thread has been going nowhere as far as I can tell, that's where. How long has it been since anything actually new has come up here?
SPOILER:
Well, tom4040 has finally realized he was wrong on the frequency change of the reflected wave, so there is some tiny progress.
#581
06-13-2012, 10:40 PM
 I_Know_Nothing Charter Member Join Date: Feb 2003 Location: Chicagoland(West Burbs) Posts: 739
Quote:
 Originally Posted by Senegoid I'm no physicist, but I drop in on this thread once every few months or so just to see where it's going. May I threadshit? I'll put it in a spoiler for those who don't want to see it. SPOILER: This thread has been going nowhere as far as I can tell, that's where. How long has it been since anything actually new has come up here? tomh4040 and others (ZenBeam, Trinopus, et many al) have simply been orbiting each other. If I measured the distance from tomh4040's arguments to any of the others today, and measured again a day, week, or year from now, will there be any change in their relative distances from one another? Do any of the arguments and rebuttals, for the last umpty posts, have any relative velocity to one another? I can't help but wonder why y'all don't just let it rest. It's not critically urgent for everybody in the world to understand relativity. I don't very much. London and Bath will still keep orbiting around their axis no matter what tomh4040's time-invariant beliefs are. Why is everybody investing all this effort to fix that? Just wond'rin.
SPOILER:
I disagree. He'll never be convinced but it does end up being a good physics lesson for those who follow along.
#582
06-14-2012, 04:33 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by I_Know_Nothing SPOILER: I disagree. He'll never be convinced but it does end up being a good physics lesson for those who follow along.
SPOILER:
Indeed. I've learned a lot trying to find ways to explain these problems to myself and others. Someone with a preconceived notion they're not willing to let go of might not learn anything, but many of the rest of us have.
#583
06-14-2012, 05:27 PM
 tomh4040 Guest Join Date: Oct 2011
You all seem to think that I had made up my mind long ago about everything that is going on in this forum. That is not so. When I started this discussion (or perhaps joined in), I did have one basic starting point, and that is that relativity is wrong, and local aether theory (or speed of light relative to the local gravitational field theory) is probably correct. I have not changed my mind on that, but through this discussion I have enlightened myself, and some of you, as to what actually is going on on our planet, and some of the ramifications of SRT.
For instance, I have not "finally realised I was wrong on frequency change with change of speed" Which Zenbeam posted. The wording "finally realised" is totally wrong and infers that this is a long held belief, which it is not. I opened up that subject not knowing any answers as I had never before asked the question. I had just assumed, as I suspect most of you had, that the speed of light in our atmosphere was c/n regardless of direction. Because of working through (sometimes tortuously) the questions and answers, and digging deeper than most people would do, and yes - playing Devil's Advocate, unforeseen answers came to the fore.

As recently as May 17th, Trinopus was still of the opinion that the speed of light here on Earth is c ( c/n ) in any direction.

Originally Posted by tomh4040
. . . If the radar increases its velocity . . .

Trinopus reply "Isn't this the core of the issue? According to SR, this doesn't happen. It's also never been observed to happen."

Have you learned something from this exchange Trinopus?
#584
06-14-2012, 05:38 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 6,893
Quote:
 Originally Posted by tomh4040 For instance, I have not "finally realised I was wrong on frequency change with change of speed" Which Zenbeam posted. The wording "finally realised" is totally wrong and infers that this is a long held belief, which it is not.
It took over two months and who knows how many posts for what's a very basic relationship. I stand by "finally realized".
#585
06-14-2012, 09:58 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by tomh4040 . . . Originally Posted by tomh4040 . . . If the radar increases its velocity . . . Trinopus reply "Isn't this the core of the issue? According to SR, this doesn't happen. It's also never been observed to happen." Have you learned something from this exchange Trinopus?
Um... No, sir. I stand by what I wrote. Can you please explain your reasons for disagreeing in precise terms? Because, as far as I know, the radar beam does not increase in velocity. SR says it doesn't, and observation does not show it to.

(Yes, I know, it changes velocity, by changing direction. But it doesn't increase velocity. That implies an increase in the absolute value of the magnitude of the vector.)

On the Straight Dope, asking people, "Do you see my point now?" has been shown to be a really, really rotten debate technique. It's your duty to make your point clear, and to defend it when it is questioned.
#586
06-24-2012, 11:36 AM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by Trinopus Um... No, sir. I stand by what I wrote. Can you please explain your reasons for disagreeing in precise terms? Because, as far as I know, the radar beam does not increase in velocity. SR says it doesn't, and observation does not show it to. (Yes, I know, it changes velocity, by changing direction. But it doesn't increase velocity. That implies an increase in the absolute value of the magnitude of the vector.) On the Straight Dope, asking people, "Do you see my point now?" has been shown to be a really, really rotten debate technique. It's your duty to make your point clear, and to defend it when it is questioned.
Trinopus, it is your duty to read the posts, and not to make (deliberate) inaccurate statements such as the one above attributed to me. I have never said that. I have answered this point quite clearly many times on this forum. For your benefit, as you have apparently missed it, I will answer it again, but if you care to look back over my posts you will find the explanation there. Bear in mind that this is the relativists explanation, not mine.
SRT says that the speed of light is c relative to any and all IFRs. The Earth is not an IFR, it is an RFR. That is an acronym which I have coined to differentiate it from other non IFRs such as a linearly accelerated frame of reference or an AFR. Restricting our discussion to east west travel at the equator, in order for the speed of light to be c relative to an IFR, it cannot be c relative to the Earth, as the Earth is an RFR. Because of the rotation of the Earth, a light beam split and sent round the Earth in opposite directions will not arrive back at the starting (entry/exit) point together. The beam going west to east will have further to go than its counterpart sent east to west and therefore take longer. This is because relative to an IFR, the Earth has rotated while the beams were circumnavigating, so shortening the east to west path and lengthening the west to east path. That is how a ring laser works.
That description is when looking from an IFR. Looking at what is happening from here on Earth, and making all measurements relative to the Earth, the beam of light traveling west to east has exactly the same distance to travel as the beam of light traveling east to west. As the east to west beam arrives back at the entry/exit point before the west to east beam, its speed was faster. There are therefore two speeds for light, with the caveat that we restrict our discussion to the equator.
If you are still of the opinion that the speed of light is c on the Earth, then by definition it is c in an RFR. If light is c in one RFR, it must be c in all RFRs. It cannot be c in a rotating ring laser (which is an RFR while it is rotating), or the two beams would arrive back at the entry/exit point together whether the device was stationary or rotating. They arrive back together if the device is not rotating (when it is an IFR), they do not arrive back together if it is rotating (when it is an RFR).
This is covered in the Wikipedia article already referenced on the subject, and is also agreed by your fellow relativists here on this forum, so you really are alone in this.
Wiki quote “If a number of stations situated on the equator relay pulses to one another, will the time-keeping still match after the relay has circumnavigated the globe? One condition for handling the relay correctly is that the time it takes the signal to travel from one station to the next is taken into account each time. On a non-rotating planet that ensures fidelity: two time-disseminating relays, going full circle in opposite directions around the globe, will arrive at the originating station simultaneously. However, on a rotating planet, it must also be taken into account that the receiver moves during the transit time of the signal, shortening or lengthening the transit time compared to what it would be in the situation of a non-rotating planet.”
From a fellow relativist :-
Quote:
 Originally Posted by ZenBeam You are correct in an inertial reference frame. In the rotating reference frame where the Earth is stationary, the E to W and W to E speeds will be different, leading to the Sagnac effect in a loop.
#587
06-27-2012, 03:23 AM
 thinkoutsidethe Guest Join Date: Jun 2012
...

I find some fundamental flaws in thought processes in this thread. If anything relativity really shows the problems of scientific observation than it does about reality.

In the example of a person traveling to a near star system at the speed of light and returning. The person will return as exactly aged as those that stayed on earth.

However, on earth they could show a second instance of the astronaut to himself through a telescope at a younger age. He would be observing himself in space time and mathematically that is what a scientist would get on his calculations.

The light reflected from the astronaut is just a hologram. Unfortunately Physics needs two things in which to observe events; matter as in machinery in which to collate information and light in which to observe (anything at a distance anyway) It cannot observe events other than that and is therefore limited in viewing reality.

Any instance in the universe can be observed by infinite amounts of perspectives with different results and none gathering an idea of reality as it actually happens unless it reverse calculates to compensate for distance and speed of light.

The problem is matter cannot travel at the speed of light unless it is in the form of energy which Einstein correctly postulates as E=MC2, So only light itself and its own speed can be used to correlate reality.
#588
06-27-2012, 04:23 AM
 Mijin Guest Join Date: Feb 2006
Quote:
 Originally Posted by thinkoutsidethe I find some fundamental flaws in thought processes in this thread. If anything relativity really shows the problems of scientific observation than it does about reality.
No, but it does show the difficulty of trying to explain the concept of relativity.
Often when a relativistic thought experiment is described to someone who is new to relativity, they object with something like "But that is just the light!" -- because it sounds like a sleight of hand, it sounds like we're saying because light has a transit time then that in itself is time dilation. It's not what relativity is, but that's how it often comes across.

I remember being told about someone travelling faster than c and why it implies time travel: but it was a badly described example and just sounded like someone seeing a set of events that had already happened in reverse order -- that's not time travel -- and it therefore took me a while to grok the time travel case.

But the important point is that light itself is not special here, c is. You could basically see relativity as space and time being warped to preserve the invariance of c in all reference frames.

Quote:
 In the example of a person traveling to a near star system at the speed of light and returning. The person will return as exactly aged as those that stayed on earth.
This is incorrect -- time is warped and the astronaut does actually age less.

Quote:
 However, on earth they could show a second instance of the astronaut to himself through a telescope at a younger age. He would be observing himself in space time and mathematically that is what a scientist would get on his calculations.
This is incorrect -- he was travelling at less than c and the light from any part of his journey has already reached earth before the astronaut.

Last edited by Mijin; 06-27-2012 at 04:27 AM.
#589
06-30-2012, 03:08 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by thinkoutsidethe I find some fundamental flaws in thought processes in this thread. If anything relativity really shows the problems of scientific observation than it does about reality. In the example of a person traveling to a near star system at the speed of light and returning. The person will return as exactly aged as those that stayed on earth. However, on earth they could show a second instance of the astronaut to himself through a telescope at a younger age. He would be observing himself in space time and mathematically that is what a scientist would get on his calculations. The light reflected from the astronaut is just a hologram. Unfortunately Physics needs two things in which to observe events; matter as in machinery in which to collate information and light in which to observe (anything at a distance anyway) It cannot observe events other than that and is therefore limited in viewing reality. Any instance in the universe can be observed by infinite amounts of perspectives with different results and none gathering an idea of reality as it actually happens unless it reverse calculates to compensate for distance and speed of light. The problem is matter cannot travel at the speed of light unless it is in the form of energy which Einstein correctly postulates as E=MC2, So only light itself and its own speed can be used to correlate reality.
Well said. Welcome to the forum. I am a heretic and a renegade, and so far, all the venom of the relativists has been aimed at me. I see you are a heretic and possibly a renegade also, so you may well attract some venom. There seems to be plenty to go round!
#590
07-01-2012, 01:38 AM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by tomh4040 Trinopus, it is your duty to read the posts, and not to make (deliberate) inaccurate statements such as the one above attributed to me. I have never said that.
I have never deliberately misattributed anything to you or to anyone else. If I have been in error, it has been honest error, and NEVER malicious disregard for the truth.

Quote:
 I have answered this point quite clearly many times on this forum.
You have not been sufficiently clear.

Quote:
 The Earth is not an IFR, it is an RFR. That is an acronym which I have coined to differentiate it from other non IFRs such as a linearly accelerated frame of reference or an AFR. Restricting our discussion to east west travel at the equator, in order for the speed of light to be c relative to an IFR, it cannot be c relative to the Earth, as the Earth is an RFR.
The Sagnac effect applies to completed paths around a ring. Not merely to light going east or west, but to light that goes around a closed path, in a ring.

Light that simply goes east, then west again, is not affected by this effect. The Michelson Morley experiment does not show any difference in speed of light beams that go east then west again, as compared to light that goes west, then east again, or north, then south.

The Earth is both an RFR and an IFR, depending on what you happen to be measuring.

Quote:
 Because of the rotation of the Earth, a light beam split and sent round the Earth in opposite directions will not arrive back at the starting (entry/exit) point together. The beam going west to east will have further to go than its counterpart sent east to west and therefore take longer. This is because relative to an IFR, the Earth has rotated while the beams were circumnavigating, so shortening the east to west path and lengthening the west to east path. That is how a ring laser works.
Yes. When the light goes in a complete ring.

This does not apply to Earth as an IFR. Light that is sent out in a straight line, and reflected back again in the same line, moves at exactly c. This supports Special Relativity.

We are in agreement. The M/M experiment, conducted in an IFR, supports the Lorentz Transformations of Special Relativity. Time, Distance, and Energy are compressed. The moving astronaut ages more slowly than the stationary astronaut.

The effect of a ring laser does not disprove any of this.

So....what are we disagreeing about, exactly?
#591
07-01-2012, 01:55 AM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by thinkoutsidethe I find some fundamental flaws in thought processes in this thread. If anything relativity really shows the problems of scientific observation than it does about reality. In the example of a person traveling to a near star system at the speed of light and returning. The person will return as exactly aged as those that stayed on earth. However, on earth they could show a second instance of the astronaut to himself through a telescope at a younger age. He would be observing himself in space time and mathematically that is what a scientist would get on his calculations.
I don't think this is correct.

To begin with, let's have the astronaut move at .99c, rather than at the speed of light. Moving at 99% of the speed of light is possible (although hideously expensive!) for a physical object. Moving at the speed of light is believed to be impossible for a physical object.

Now: he goes to Alpha Centauri and back. Say, 4.00 light years for convenience. It takes him 4.04 years, each way. By the Lorentz equations, he only ages about half a year on each leg of the trip, so after the eight year round trip, he's only aged one year.

Now: exactly how do you propose that "they show a second instance of the astronaut to himself through a telescope at a younger age." I don't understand what you mean by this.

Once he has returned, he isn't in space any longer, and so they can't "show" anything to him. He's here.

When he is in space, yes, he can send pictures of himself back to earth. People on earth can send him pictures. They can send him pictures of New Year's Eve, 2013, 2014, 2015, 2016, etc. He is amazed, as these "years" seem to him to be a month and a half apart.

He can't send a picture of himself to himself. The light waves of the tv transmission move faster than he is moving.

He could send a messenger rocket to himself, but this would add additional complications; you'd have to calculate the speed of this messenger, and how it related to all the other frames of reference. It's pointlessly messy.

Quote:
 The light reflected from the astronaut is just a hologram.
Not a hologram, exactly, but let's simply say a light-image. This is well understood. When we look at Alpha Centauri, we're seeing light that is four years old. We're seeing the light-image of AC. We can't see AC as it is "right now," because it takes four years for the light to get here.

Quote:
 Unfortunately Physics needs two things in which to observe events; matter as in machinery in which to collate information and light in which to observe (anything at a distance anyway) It cannot observe events other than that and is therefore limited in viewing reality.
Generally true. Alpha Centauri might have gone supernova three years ago, but we wouldn't know it for another year.

What Relativity tells us is that events that may seem simultaneous in one frame of reference may seen non-simultaneous to another frame of reference.

Quote:
 Any instance in the universe can be observed by infinite amounts of perspectives with different results and none gathering an idea of reality as it actually happens unless it reverse calculates to compensate for distance and speed of light. The problem is matter cannot travel at the speed of light unless it is in the form of energy which Einstein correctly postulates as E=MC2, So only light itself and its own speed can be used to correlate reality.
Again, I'm not totally clear what you mean, but I think I agree. We have to calculate events based on the relative speeds of various frames of reference.

You and I might be in the same place at a specific instant, but you're stationary, and I'm moving very, very fast. We're both watching two distant stars, which go supernova. You might see them both go supernova at the same instant; yet I might see one go supernova before the other one does.

Neither of us is "wrong." Both of us are right. You see two events as simultaneous; I see them as sequential. That's the price of relativity.
#592
07-01-2012, 02:01 AM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by Mijin . . . I remember being told about someone travelling faster than c and why it implies time travel: but it was a badly described example and just sounded like someone seeing a set of events that had already happened in reverse order -- that's not time travel -- and it therefore took me a while to grok the time travel case. . . .
My old physics prof said that if you could travel FTL, or even if you could send a message FTL, then you could travel or send a message backward in time.

I've never quite managed to understand this. I don't know if it has been demonstrated using proper mathematics, or if it is an appeal to the unknown, since there is no physical meaning to the "square root of a negative number." I've done a lot of Google searching on this, and haven't seen a solid explanation.

(Meanwhile, you do have the possibility of time travel in Tipler machines and other extravagances. Fortunately, no one is likely to send a probe in a high-speed closed orbit around a gigantic, hyper-massive cylinder that is spinning at extremely high rates of revolution! Tipler, Penrose, Hawking, and others suggest that this could lead to time travel...but I think the experiment will never actually be performed!)
#593
07-01-2012, 07:38 AM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 6,893
Quote:
 Originally Posted by Trinopus My old physics prof said that if you could travel FTL, or even if you could send a message FTL, then you could travel or send a message backward in time. I've never quite managed to understand this. I don't know if it has been demonstrated using proper mathematics, or if it is an appeal to the unknown, since there is no physical meaning to the "square root of a negative number." I've done a lot of Google searching on this, and haven't seen a solid explanation.
This argument comes about because frames of reference moving with respect to each other don't agree on what planes of constant time would be. If you look at the Minkowski diagram at the top of that page, t = 0 in the black reference frame doesn't agree with t' = 0 in the blue frame.

To make it easier, imagine a signal can be sent instantaneously. In the black frame, send a signal to the right. Now accelerate to the right until you're in the blue frame, and send a signal instantaneously to the left in that frame. It can get back to x = 0 in the black frame in the past, in that frame.

If the signals could be sent FTL, but not instantaneously, that may mean you need to accelerate a lot to send signals into the past.

It's not strictly true that FTL signals always allows sending a signal back in time. It requires there also to be no privileged reference frame*. For example, if the black frame were a privileged frame, and these FTL signals were always instantaneous in that frame no matter what frame the sender was in, you wouldn't be able send signals backwards in time.

* Our physics doesn't have privileged frames, but it also doesn't have FTL signals.

Last edited by ZenBeam; 07-01-2012 at 07:39 AM.
#594
07-02-2012, 06:16 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by ZenBeam This argument comes about because frames of reference moving with respect to each other don't agree on what planes of constant time would be. If you look at the Minkowski diagram at the top of that page, t = 0 in the black reference frame doesn't agree with t' = 0 in the blue frame. . . .
I'm embarrassed and ashamed...but honest enough...to admit I don't understand this. Until this thread, I'd never heard of the Sagnac Effect, either.

(I'm only participating in this thread because questions have been asked which I *do* know the answers to!)

Oh, well... I guess I'll just brew up a nice cup of Thiotimoline tea...
#595
07-03-2012, 06:09 AM
 Half Man Half Wit Guest Join Date: Jun 2007
Two spaceships zip past each other at 87%c relative speed. Ten minutes after, you (on spaceship A) decide to send an (instantaneous) message to spaceship B, because their left rear light is broken. On spaceship B, from your point of view, five minutes have elapsed since the meeting, due to time dilation (at 87%c, the Lorentz factor is about 2). So the instantaneous message arrives five minutes after the meeting on B. The return message is sent immediately ('Thx, fixed!'), and, since from their point of view, only 2.5 minutes have elapsed on your ship (since you are moving, relative to them, at 87%c and five minutes have elapsed on their on-board clock), the instantaneous message reaches you 2.5 minutes after the meeting, and 7.5 minutes before you ever sent the original message in the first place. Which you now don't need to, seeing as how they've long fixed their rear light.

#596
07-03-2012, 04:33 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by Trinopus I have never deliberately misattributed anything to you or to anyone else. If I have been in error, it has been honest error, and NEVER malicious disregard for the truth. You have not been sufficiently clear. The Sagnac effect applies to completed paths around a ring. Not merely to light going east or west, but to light that goes around a closed path, in a ring.
You attributed this quote to me, but you edited a pertinent part of it out. Here is the quote again with the second part in place :-
"Because of the rotation of the Earth, a light beam split and sent round the Earth in opposite directions will not arrive back at the starting (entry/exit) point together. The beam going west to east will have further to go than its counterpart sent east to west and therefore take longer. This is because relative to an IFR, the Earth has rotated while the beams were circumnavigating, so shortening the east to west path and lengthening the west to east path. That is how a ring laser works." That was what you quoted and is OK so far. This is what you omitted, did you agree to this?
"That description is when looking from an IFR. Looking at what is happening from here on Earth, and making all measurements relative to the Earth, the beam of light traveling west to east has exactly the same distance to travel as the beam of light traveling east to west. As the east to west beam arrives back at the entry/exit point before the west to east beam, its speed was faster. There are therefore two speeds for light, with the caveat that we restrict our discussion to the equator."
Quote:
 Yes. When the light goes in a complete ring.
You are forgetting a very important point here. Light always travels in straight lines (ignoring geodesics which bend the path of light very gradually and by a very small amount, and are not applicable here on Earth). When light is sent round the Earth, it uses mirrors or fibre optics. As it travels round the Earth it goes in a straight line, it is then reflected through a small angle by a mirror (or the inside of the fibre optic), then in a straight line, then reflected again until it arrives back at its starting point. In other words it always travels in a straight line. Let us assume that it travels 400 legs in straight lines, being reflected by mirrors in between each leg. Each leg is then 100 kilometers long. As the speed difference is detected at the end of the journey when the light is back at its entry/exit point, there must be a speed difference when light travels in a straight line. If there is a speed difference detected at the end of the loop, precisely when does that speed difference appear? Does it increase in 400 small increments or does it appear in just one of the 400 legs – if so, which one? No, the answer is that light, no matter how short its journey and whether it is reflected or not, has the speeds (at the equator) :-
Velocity of light west to east (VLWE) = 299,792,019.2 M/s
Velocity of light east to west (VLEW) = 299,792,946.8 M/s

Quote:
 Light that simply goes east, then west again, is not affected by this effect. The Michelson Morley experiment does not show any difference in speed of light beams that go east then west again, as compared to light that goes west, then east again, or north, then south.
The MMX does not detect a difference because it is measuring the average speed of light. The average speed east to west or west to east is the same as the average north to south or south to north.

Quote:
 The Earth is both an RFR and an IFR, depending on what you happen to be measuring.
This is preposterous, and leads to the conclusion that if an RFR can be an IFR, then a ring laser can be an IFR, and you only included it to sidestep my point which was :- “If you are still of the opinion that the speed of light is c on the Earth, then by definition it is c in an RFR. If light is c in one RFR, it must be c in all RFRs. It cannot be c in a rotating ring laser (which is an RFR while it is rotating), or the two beams would arrive back at the entry/exit point together whether the device was stationary or rotating. They arrive back together if the device is not rotating (when it is an IFR), they do not arrive back together if it is rotating (when it is an RFR).”
Please answer this question. We know that in a (rotating) ring laser the two beams have different speeds or they would arrive back together. What size does a ring laser have to be before it ceases to be an RFR and becomes an IFR so that the two beams of light sent round in opposite directions arrive back at the entry/exit point together?

Quote:
 This does not apply to Earth as an IFR. Light that is sent out in a straight line, and reflected back again in the same line, moves at exactly c. This supports Special Relativity.
The Earth is not an IFR, and the speed of light is c only in an IFR. This point has already been covered. Light going round the Earth goes in straight lines, and during each leg its speed is less than or more than c depending on its direction. If radar did not behave the same, a radar pulse sent between two mirrors alongside the light beam would not have the same speed as the light beam. This is definitely against relativity.

Quote:
 We are in agreement. The M/M experiment, conducted in an IFR, supports the Lorentz Transformations of Special Relativity. Time, Distance, and Energy are compressed. The moving astronaut ages more slowly than the stationary astronaut.
We are not in agreement. The MMX cannot differentiate between aether theory, or light speed relative to the source, or SRT, and it was not, and any repeats of it are not, conducted in an IFR, they are conducted in an RFR. Neither does it support time dilation et al, they are suppositions of SRT which are borne out by the mathematics of the Lorentz transformations. This is not a coincidence, SRT was invented to match the Lorentz transformations.

Quote:
 The effect of a ring laser does not disprove any of this.
The ring laser proves that the speed of light is not c in an RFR. All IFRs are equal, and all RFRs are equal, so must be treated as such. The Earth is an RFR, so the speed of light (round the equator) is not c relative to it.

Quote:
 So....what are we disagreeing about, exactly?
We are disagreeing about all the above points. Remember that I am playing the part of a relativist in this discussion.
#597
07-03-2012, 05:43 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 4,784
Quote:
 Originally Posted by tomh4040 You attributed this quote to me, but you edited a pertinent part of it out. . . .
I do not play these games.

Ask a question, and the participants here will try to answer it.
#598
07-04-2012, 04:26 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by Trinopus I do not play these games. Ask a question, and the participants here will try to answer it.
You certainly do play games. You pretend to know what you are talking about, then admit on this forum that you had never heard of the Sagnac effect before it was discussed here.
Like John Cazale, you ask questions of others, then refuse to answer ones put directly to you.

 Bookmarks

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is Off HTML code is Off
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Main     About This Message Board     Comments on Cecil's Columns/Staff Reports     Straight Dope Chicago     General Questions     Great Debates     Elections     Cafe Society     The Game Room     In My Humble Opinion (IMHO)     Mundane Pointless Stuff I Must Share (MPSIMS)     Marketplace     The BBQ Pit Side Conversations     The Barn House

All times are GMT -5. The time now is 03:18 AM.