# B Effect of Varying Distance on Radio Waves

1. May 22, 2017

### RAMII19780529

Hello -

Not sure if this is the correct location for this.

I've been thinking about how a radio wave would work as the transmitter travels further away from the receiver.

My example is a transmitter on a spaceship traveling 60,000 kph that transmits a loop of my favorite playlist back to earth.

After ~18,000 hours, the spaceship will be about a light-hour away.

What would I hear on earth? Does the transmission sound 0.0055% slower on earth to compensate for the speed of the spaceship, so when it's streaming from 1 light-hour away, the delay has already been included in the playback?

Would this also change the radio frequency I would need to "tune-in" to a lower frequency than what's being transmitted based on how fast the spaceship is moving?

Thanks!

2. May 22, 2017

### Staff: Mentor

It depends on the design of the receiver. What is being transmitted through space isn't sound; the radio waves have to be converted to sound in the receiver, and that process is not as straightforward as you appear to be assuming.

This question does have a simple straightforward answer: yes.

3. May 22, 2017

### RAMII19780529

For clarification, lets keep it simple and say we are talking about an analog signal such as AM where the difference in amplitude is used to carry the signal.

4. May 22, 2017

### Staff: Mentor

That still doesn't necessarily mean the signal carried in the radio wave is identical to the signal output as sound at the speakers. In fact, a big part of the design of radio receivers is to make that not true, by compensating for all kinds of variations in the input signal, including variations in the frequency.

If for purposes of discussion we assume a really primitive receiver that doesn't have all that fancy compensation in it (something like an early crystal radio set), then there would be two effects: the sounds would be slightly lower in pitch when heard on Earth than on the spaceship, and it would take slightly longer for the loop to complete than it does on the spaceship.

Btw, the effect here is actually a combination of two things: the spaceship moving away, which means the travel time of the signal from spaceship to Earth is continually increasing; and relativistic time dilation. The fully relativistic formula for the frequency received in terms of the frequency emitted is

$$f_r = f_e \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}}$$

For $v = 60000$ kph, this works out to $f_r / f_e = 0.999967$, or a shift of about $0.0033$ percent.

5. May 22, 2017

### RAMII19780529

I really wasn't concerned with the technology of the transmitter and receiver, or whatever type of filters they may used to "improve" the sound quality. I used that as an example to help me get my head around what is happening.

I was pretty sure that both the speed and distance need to be taken into consideration, but the lower pitch / slower playback was related to the speed, not the distance, as I assume that if the spaceship is standing still in relation to me, then the sound would play back exactly as it does here, except it would be delayed by 1 hour. That one hour must go somewhere when we are listening to a constant stream over the distance, so I assumed that it would be received slightly slower while the spaceship was moving away.

I'm curious about the math. I got 0.0055% by following this train of thought: one light-hour is about 1,080,000,000 km. so when the spaceship hits that mark, the stream must have been delayed by 1 hour at the receiver. Since it takes 18,000 hours, traveling at 60,000 kph, the amount of time lost would have been 1/18,001 which is 5.55e-5, which is 0.0055%.

It seems to work out correctly when making a change to the distance as well... calculating a 24 hour loss from a light-day away going 60,000 kph sill gives me 0.0055%. Only changing the speed makes a difference.

What did I do wrong? Am I making an incorrect assumption somewhere here?

Thank you for taking the time to help me figure this out!

EDIT:
If I use the equation you gave me, I too get 0.0033% but the problem with that is when I use it to calculate the actual time loss, it says I would have lost less time than I actually did.

Last edited: May 22, 2017
6. May 22, 2017

### Staff: Mentor

Delayed in the sense that, in the common rest frame of you and the spaceship, it takes an hour for the signal to get from the spaceship to you, yes.

The effects I described don't depend on how far away the spaceship is from you; they only depend on its speed away from you. The only thing the distance affects is the propagation delay; but, as the example of the ship standing still shows, the propagation delay cancels out. What doesn't cancel out is the change in the propagation delay due to the ship's motion relative to you. (And also, if we take relativity into account, the time dilation of the ship relative to you.)

You seem to be assuming that the ship's distance from you makes a difference to the observed frequency shift. It doesn't; see above. Also, you're not taking into account time dilation.

For a simple derivation of the correct relativistic Doppler formula, see here:

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight

This derivation automatically takes into account both effects that I mentioned: the ship's speed away from you and the corresponding change in propagation delay, and the ship's time dilation relative to you.

7. May 23, 2017

### RAMII19780529

I don't make that assumption. I said that the distance doesn't make a difference, only the speed does.

I'm familiar with basic concept of the Doppler effect when talking about sound waves, but I didn't know the equations. Thanks for that, I'll have to look into those a bit more.

I think this is what I'm missing! I have no concept of time dilation. I never learned this in school. From earth, time would appear to pass slower on the spaceship, and from the spaceship, time would appear to pass slower on earth? This is getting into uncharted territory in my mind. I don't understand how any of that applies, and to be honest, the special relativity equations are currently over my head.

I guess I can see where I am bringing the distance into this... This is the how I look at this problem, without equations, just thinking about it, all with a perspective from earth...

When the spaceship is next to me on the ground, the radio is playing with basically no delay due to distance, and no frequency distortion due to relative speed (zero).

When the spaceship is 1,080,000,000 km away from me, the signal that leaves the ship would take 1 hour to reach me on earth.

I think this is where my mistake is: After traveling 60,000 kph for 18,000 earth hours (when the spaceship reaches a distance of 1,080,000,000 km) it stops transmitting. I would have only heard 17,999 hours of music since the music currently leaving the ship at the 18,000 hour mark will take 1 hour to get back to earth based on it's distance. If the spaceship stopped transmitting after exactly 18,000 earth hours, the last signal would be received on earth after 18,001 hours. Therefor I would have heard 18,000 hours of music in 18,001 hours, which gives me the 0.0055% I calculated. But I think this is wrong now...

If time is moving slower on the spaceship due to special relativity, then I guess the full 18,000 hours worth of music would not have been played in 18,000 earth hours. Using the equation you provided, I calculate that it would have transmitted 17,999.4 hours of music instead, which would still take 18,001 hours for the end of the signal to reach earth. I'm so confused.

I'm now looking to figure out a few things. After 18,000 earth hours, how many hours of music would I have been transmitted from the spaceship and how long will it take me to hear it all? I came up with 17,999.4 hours of music in 18,001 hours. But I think I am using the wrong units somewhere here.

Thanks!

EDIT:
I don't think I did any of the calculations correctly. I feel so dumb... I really want to understand this stuff.

Last edited: May 23, 2017
8. May 23, 2017

### PeroK

You posted this in the relativity forum which is why you got an answer involving the relativistic Doppler shift.

The speed of your spacecraft is hardly relativistic, so you could equally use the classical Doppler shift and get effectively the same answer.

9. May 23, 2017

### PeroK

PS I get 0.0055% for 60,000 kph using both formulas, classical and relativistic. The relativistic formula reduces to $1 - v/c$ to first order in $v/c$.

10. May 23, 2017

### RAMII19780529

I asked this in the relativity forum because of the extra time it would take for the radio signal to reach me when the space ship was 1 light-hour away, but I don't think this was the correct location now. I also believe I understand what I did wrong here. Below is how I calculated the Doppler shift using both the standard and the relative equations. They are very close, because as you said, 60,000 kph is really slow as to not make a big difference. Based on this thread, I now have a better understanding, but I'm still not sure of how the distance comes into this when discussing the actual length of time the music would play v.s. the length of music I would actually hear if the spaceship stopped transmitting after 18,000 hours from earths perspective. I'm not even sure if I'm asking that correctly.

v = 60,000 kph / 60 = 1,000 kps
c = 300,000 kps

v/c
1,000 / 300,000 = 1 / 300 = 0.003333

((1 - v/c) / (1 + v/c) ) ^ (1/2)
1 - 0.003333 = 0.996666
1 + 0.003333 = 1.003333
0.996666 / 1.003333 = 0.993355
1 - (0.993355 ^ (0.5)) = 0.003327

If I take the difference between those, I get 0.000006, which I assume is the amount related to the relativity. So if I take that and multiply it by the number of hours it was transmitting, I get 6 minutes 29 seconds. So I'm guessing that is the amount of music that would be missing when the spaceship stopped transmitting.

Last edited: May 23, 2017
11. May 23, 2017

### PeroK

Unfortunately, I think you'll find there are 3600 seconds in an hour!

12. May 23, 2017

### RAMII19780529

LOL OMG... Thanks!

I corrected my equations and I'm getting 5.5554x10-5 and 5.5555x10-5 so both equation are basically the same. Fractions of a second in difference. Which brings me back to my original question, which I can answer correctly. It would take 18,001 hours to listen to 18,000 hours and the math matches! Ignoring the fraction of a second that was lost.