1. Sep 4, 2010

### Passionflower

It is very easy to establish distance and velocity by measuring the roundtrip time of a radar signal that is reflected by a moving observer. However does anybody have the formulas to determine the acceleration which should be describable in terms of an increase or decrease of the roundtrip rate.

For example assume the roundtrip rate = 4, this means the observer is moving with a velocity of 0.6. Now the observer starts to accelerate, how do we express the acceleration (either proper or coordinate) in terms of an increase of the rate?

2. Sep 4, 2010

### bcrowell

Staff Emeritus
http://www.phys.uu.nl/igg/dieks/rotation.pdf [Broken]

See p. 9.

Last edited by a moderator: May 4, 2017
3. Sep 4, 2010

### Passionflower

Thank you for the reference, it is not quite what I am looking for but it does highlight a yet different interesting aspect as well.

In the reference you gave Prof. Dieks talks about the radar roundtrip time between two locations in an accelerating frame of reference from the perspective of an inertial observer which is not quite what I am looking for but interesting nevertheless.

This is what I am looking for:

If we have two inertial observers O1 and O2 measuring radar roundtrip times T1 and T2 between them how do these times relate to relative acceleration and proper acceleration?

For O1:

If dT= 0 then v = 0 and a = 0.
If dT ≠ 0 and d2T = 0 then v = (dT-1)/(dT+1) and a = 0
If dT ≠ 0 and d2T ≠ 0 then how do we express a?

The acceleration a for O1 is obviously not necessarily equal for O2 as it depends on the direction and who is accelerating. Also either O1 or O2 is accelerating or both are and this seems to have an effect on T for both observers as well.

So anybody ever got this worked out?

Last edited by a moderator: May 4, 2017
4. Sep 4, 2010

### bcrowell

Staff Emeritus
I'm pretty sure no such relationship exists. When you talk about inertial observers who are accelerating relative to one another, that means you're talking about curved spacetime. The round trip times could, for example, be infinite if the radar beams were in circular orbits around a black hole.

Re proper acceleration, shouldn't the proper accelerations be zero, since they're inertial observers?

5. Sep 5, 2010

### Passionflower

Oh dear I seem not the be very successful in explaining this.

I meant that at one point if one or both will start to accelerate how does that influence the radar roundtrip time?

Perhaps I should make a headstart and then someone can pitch in later:

Let's assume A and B are inertial, at t=0 B moves away with a constant proper acceleration $\alpha$. Can A determine $\alpha$ by measuring the change in radar roundtrip rate?

In the travels with a constant velocity v, the radar roundtrip rate is constant:

If we relativistically double the speed (double the rapidity) then the new rate becomes r2. So it appears we are looking for some exponential function.

Last edited: Sep 5, 2010
6. Sep 5, 2010

### marcusl

The roundtrip echo delay time gives you range only. Relative radial velocity is most commonly measured by the Doppler shift.

You could instead look at two pulses and calculate $$\Delta R / \Delta t$$. Three pulses will give an estimate of acceleration.

7. Sep 5, 2010

### Passionflower

Obviously the roundtrip rate change can only be measured by at least two signals.
But hat do you mean by "range"?

Ok, so what is the formula?

8. Sep 5, 2010

### bcrowell

Staff Emeritus
If B is inertial, isn't his proper acceleration zero?

9. Sep 5, 2010

### Passionflower

Initially A and B are inertial, then at t=0 B a moves away with a proper constant acceleration.

10. Sep 5, 2010

### bcrowell

Staff Emeritus
In that case, let t be measured on A's clock, and let D be the round-trip time as measured on A's clock. A infers B's position to be D/2 (in units with c=1), and his acceleration to be $d^2(D/2)/dt^2$. (In practice, D is measured as a discrete set of points, not a smooth curve, so A has to use differences to estimate the second derivative.)

11. Sep 5, 2010

### marcusl

Range is radar-speak for distance to the target. Why don't you find the (non-relativistic) formula for acceleration? Hint: use the fundamental definition of derivative.

12. Sep 5, 2010

### Passionflower

If you don't have it or don't want to give it fine, but that is what I am asking for the formula. These things are not supposed to be secrets.

With regard to range, by emitting consecutive signals one can determine based on the rate of increase (or decrease) in time the velocity of the target. This rate is the square of the Doppler factor. For a constant velocity roundtrip time increases monotonically, however for an accelerating target the rate increases exponentially but quickly approaches a monotonic increase due to relativity.

I seem to miss something here.

So a concrete example:

Let's say that at t=1 the roundtrip time is 1.441518, at t=2 the roundtrip time is 3.388508 and at t=3 the roundtrip time is 5.370257 then by using your formula what is the acceleration?

Last edited: Sep 6, 2010
13. Sep 6, 2010

### JesseM

In an edit to post #13 on this thread I showed that if a reflective surface is moving towards you with speed v, and you are sending signals with a frequency of fsent and they are coming back to you at a frequency of freceived, then the relationship between the two (same in both SR and Newtonian physics) is:

$$f_{received} = f_{sent} \frac{1 + v/c}{1 - v/c}$$

Since frequency is just one over the period (i.e. the time interval between successive signals), this is equivalent to:

$$P_{received} = P_{sent} \frac{1 - v/c}{1 + v/c}$$

So abbreviating Preceived as Pr and Psent as Ps, we can solve this for v:

Pr*(1 + v/c) = Ps*(1 - v/c)
Pr - Ps = -v/c*(Pr + Ps)
$$v = c \frac{P_s - P_r}{P_s + P_r}$$

So if you send out two signals 1 and 2 with a time interval of Ps between them, and they come back to you with a time interval of Pr, you can conclude that even if the reflective object was accelerating, its average velocity between the moment signal 1 bounced off it and the moment signal 2 bounced off it must have been $$c \frac{P_s - P_r}{P_s + P_r}$$. Then if you send another signal such that there was again a time interval of Ps between 2 and 3, and it comes back to you a time interval of P'r after signal 2 came back to you, you can conclude the average velocity of the reflective object between the moment signal 2 bounced off it and the moment signal 3 bounced off it must have been $$c \frac{P_s - P'_r}{P_s + P'_r}$$.

You also know that the time between signal 1 bouncing off the object and signal 2 bouncing off the object was Pr, and the time between signal 2 bouncing off the object and signal 3 bouncing off the object was P'r. So, if we pick two events at the midpoint of the time intervals between successive bounces, the time between them must be (Pr + P'r)/2. So, if we idealize that the average velocity between bounces was also the exact velocity at the midpoints of the time intervals between bounces--which I think would be a reasonable idealization in the limit as these intervals became shorter and shorter--then it seems like (change in velocity)/(change in time) should be:

$$2c( \frac{P_s - P'_r}{P_s + P'_r} - \frac{P_s - P_r}{P_s + P_r}) / (P_r + P'_r)$$

There may well be an error in my reasoning here though, if anyone sees one please point it out...

14. Sep 6, 2010

### Passionflower

JesseM and bcrowell I appreciate you are helping to solve this problem (there are three scenarios this is just the simplest part, the other is the radar time from the perspective of the accelerating observer and the situation where both accelerate away from each other).

I checked the literature and simply cannot find the formula. Now surely that must be my problem but I searched pretty hard. :)

So those who know those formulas (Marcusl?) let's get these formulas out for everybody's edification!

JesseM, I follow you and we indeed got the formula relating velocity and roundtrip rate lock, stock and barrel, it is in fact the square of the Doppler shift.

Now acceleration, I follow what you say up to:
Are you targeting the proper or coordinate acceleration here? Any of them is good as we can derive one from the other easily.

If you chart the coordinate distances in coordinate time for the traveler undergoing a constant proper acceleration (see attachment 1) you can see that very quickly the relationship between coordinate time and coordinate distance approaches linearity due to the fact that the coordinate acceleration tends to 0 (see attachment 2) while the proper acceleration remains the same. It becomes rather 'dramatic' when you plot using $-\alpha$ an approaching traveler with a constant proper acceleration approaching the observer.

The distance function is here:
First get $\tau$:

$$\tau = (asinh (\alpha t)) / \alpha$$

To get the coordinate distance d:

$$d = (\cosh(\alpha \tau) - 1) / \alpha$$

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Last edited: Sep 6, 2010
15. Sep 6, 2010

### bcrowell

Staff Emeritus
The function D/2 equals 0.721, 1.694, and 2.685 at one-second intervals. Its derivative, estimated from finite differences, equals 0.973 and 0.991 at times separated by a one-second interval. The second derivative is the difference of these two, which is .018. This is what A infers B's acceleration to be.

The question seems trivial to me. The only relativity involved would be if you also wanted to transform into B's frame in order to find out what B says A's acceleration is.

16. Sep 6, 2010

### Passionflower

If so trivial you should perhaps make your own verifications.

Clearly the solution is not algebraic, to show you are wrong, here is an array of results for $\alpha=0.1$:
Code (Text):

Time    Distance    Distance/2  Velocity    Difference  Difference  Difference
1   0.049875621 0.024937811 0.099503719
2   0.198039027 0.099019514 0.196116135 0.074081703
3   0.440306509 0.220153254 0.287347886 0.121133741 0.047052038
4   0.770329614 0.385164807 0.371390676 0.165011553 0.043877812 -0.003174226
5   1.180339887 0.590169944 0.447213595 0.205005137 0.039993584 -0.003884228
6   1.66190379  0.830951895 0.514495755 0.240781951 0.035776814 -0.004216769
7   2.206555616 1.103277808 0.573462344 0.272325913 0.031543962 -0.004232853
8   2.806248475 1.403124237 0.624695048 0.29984643  0.027520517 -0.004023445
9   3.453624047 1.726812024 0.668964732 0.323687786 0.023841357 -0.00367916
10  4.142135624 2.071067812 0.707106781 0.344255788 0.020568002 -0.003273354
11  4.866068747 2.433034374 0.739940073 0.361966562 0.017710773 -0.002857229
12  5.620499352 2.810249676 0.76822128  0.377215302 0.01524874  -0.002462033
13  6.401219467 3.200609733 0.792623989 0.390360058 0.013144755 -0.002103985
14  7.204650534 3.602325267 0.813733471 0.401715534 0.011355476 -0.001789279
15  8.027756377 4.013878189 0.832050294 0.411552922 0.009837388 -0.001518088

The differences are for the distances
We can 'difference' forever but it will never be 0.

17. Sep 6, 2010

### JesseM

Yes, the square of the relativistic Doppler shift. Note, though, that the formula I gave was not specifically relativistic, it would also work in a Newtonian universe for an observer who's in a frame where waves move at the same speed c (which could just represent the speed of sound waves in some medium) in both directions.
Coordinate acceleration. Again, I'm using the approximation that the velocity at the midpoint between two signals bouncing was equal to the average velocity between the bounces, and then I'm dividing the difference in velocities by the time interval between the time at the midpoint between bounce 1 and bounce 2 and the time at the midpoint between bounce 2 and bounce 3. So, it's just change in coordinate velocity (approximate) divided by change in coordinate time.
Tends to zero, but the coordinate acceleration is always nonzero, so if my formula is right then with a sufficiently small time interval between signals one should be able to detect it. I calculated the formula for coordinate acceleration as a function of time given constant proper acceleration ap in post #143 here, it worked out to:

a(t) = ap*(1 + (apt/c)2)-3/2

18. Sep 6, 2010

### bcrowell

Staff Emeritus
Are you saying that you expect the nth difference to be identically zero for some n? Why? If B had constant acceleration in A's frame of reference, then the third derivative would be zero. But your figures are for the case where B has constant proper acceleration, not constant acceleration in A's frame.

Note that in the notation I defined in #10, D is not the distance, it's the round-trip time, which is approximately equal to twice the distance.

Other than that factor of 2, your table behaves exactly as I would expect. The column headed "Difference" and beginning with 0.047 equals half the acceleration given by the approximation in my #10. Doubling that first entry, you get about 0.094. This is close to the proper acceleration of 0.1, which makes sense because the velocity is fairly nonrelativistic at that point. Later on, that column approaches zero, which also makes sense, because B approaches the speed of light.

19. Sep 6, 2010

### Passionflower

I don't expect it but you clearly did and you, seemingly incorrectly, implied that an algebraic formula would work.

Yes, that was a given, are you now implying you were not aware of that?

So what are you implying that you were right? Clearly your approach does not work.

What we need is a formula that relates the change in roundtrip rate with the rate of acceleration, not approximately but exactly.

Since you call this a trivial problem, then what is stopping you from writing down the formula? Then I plug it into Matlab and we can verify it is correct or not.

The formula should give Alpha = 0.1 based on the following rates:

Code (Text):

Time    Distance    Roundtrip Rate
1   0.049875621 1.220997512
2   0.198039027 1.487921561
3   0.440306509 1.806418391
4   0.770329614 2.181626369
5   1.180339887 2.618033989
6   1.66190379  3.119428455
7   2.206555616 3.688917786
8   2.806248475 4.328999756
9   3.453624047 5.041652328
10  4.142135624 5.828427125
11  4.866068747 6.690535124
12  5.620499352 7.628919844
13  6.401219467 8.644317061
14  7.204650534 9.73730215
15  8.027756377 10.90832691

JesseM did you numerically verify your formula?

Last edited: Sep 6, 2010
20. Sep 6, 2010

### bcrowell

Staff Emeritus
Passionflower, you seem upset. I'm not going to put further effort into helping you if it makes you upset.