Wow. Just … amazing. You don’t even understand the relationship between period and frequency. This is so basic…
From Wikipedia:
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:
I have just got to say that the difference between RFRs and** IFR**s 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:
[ul]
[li]There is no Doppler shift observed in a ring laser in a RFR.[/li][li]There is a fringe pattern observed in a RFR.[/li][li]The RFR observer sees a difference in c.[/li][li]The IFR observer sees no difference in c.[/li][/ul]
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.
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 = vt = 2e840e-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 .
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.
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.
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.
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.
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).
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!
More evasion, Tom. Again, you said:
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.
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.
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.
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.
Spectacular! N ow, where’s the second observer?
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 rcos(v/rt) and the car’s y coordinate is rsin(v/rt) - so radial distance never changes, and thus the radial velocity is zero - but the total velocity is never zero.
Spoiler! It doesn’t.
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.
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.
Please point me to a ring laser/interferometer result which proves that light speed is not constant WRT the Earth.
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.
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.
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.
Well, tom4040 has finally realized he was wrong on the frequency change of the reflected wave, so there is some tiny progress.