How can radar data be used to calculate acceleration?

[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?

 

[bcrowell]

http://www.phys.uu.nl/igg/dieks/rotation.pdf

See p. 9.

 

[Passionflower]

http://www.phys.uu.nl/igg/dieks/rotation.pdf

See p. 9.

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?

 

[bcrowell]

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?

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?

 

[Passionflower]

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?

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:

Radar roundtrip rate = (1+v)/(1-v)

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.

 

[marcusl]

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.

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.

 

[Passionflower]

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

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

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

Ok, so what is the formula?

 

[bcrowell]

Let's assume A and B are inertial, at t=0 B moves away with a constant proper acceleration $\alpha$.

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

 

[Passionflower]

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

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

 

[bcrowell]

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

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.)

 

[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.

 

[Passionflower]

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.

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.

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.)

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?

 

[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 f_sent and they are coming back to you at a frequency of f_received, 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 P_received as P_r and P_sent as P_s, we can solve this for v:

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

So if you send out two signals 1 and 2 with a time interval of P_s between them, and they come back to you with a time interval of P_r, 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 P_s 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 P_r, 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 (P_r + 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...

 

[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:

So, if we pick two events at the midpoint of the time intervals between successive bounces, the time between them must be (P_r + 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...

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:

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

 

[bcrowell]

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?

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.

 

[Passionflower]

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.

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 $\alpha=0.1$:

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.

 

[JesseM]

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.

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.

Are you targeting the proper or coordinate acceleration here? Any of them is good as we can derive one from the other easily.

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.

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)

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 a_p in post #143 here, it worked out to:

a(t) = a_p*(1 + (a_pt/c)²)^-3/2

 

[bcrowell]

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 $\alpha=0.1$: [...]
The differences are for the distances
We can 'difference' forever but it will never be 0.

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.

 

[Passionflower]

Are you saying that you expect the nth difference to be identically zero for some n? Why?

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

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.

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

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.

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:

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?

 

[bcrowell]

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

 

[marcusl]

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.

You don't need to be grouchy. One of the purposes of Physics Forums is to educate by having posters try to figure things out for themselves. If that's not for you, fine. Here:

$$\hat{a}(t)=\frac{c[\tau(t+2T) - 2\tau(t+T) + \tau(t)]}{2T}$$

where $$\tau(t)$$ is the round trip echo delay at time t and T is the pulse repetition interval.

 

[JesseM]

JesseM did you numerically verify your formula?

Looking things over, I just realized I did make an error in my reasoning--I assumed that the time P_r between my receiving the reflections of signals 1 and 2 was equal to the time between signals 1 and 2 actually bouncing off the reflective surface, but of course since the reflective surface is moving in my frame this wouldn't be the case! If the reflective surface had (change in position)/(change in time) given by $$v = c \frac{P_s - P_r}{P_s + P_r}$$ between the two signals bouncing off it, then if T is the time between the two bounces, the change in position must have been Tc(P_s - P_r)/(P_s + P_r). So assuming without loss of generality that v is positive (meaning the surface is traveling towards me), if the two reflective signals were emitted a time T apart from one another but the second's distance to reach me was shorter than the first one's distance by Tc(P_s - P_r)/(P_s + P_r), then the time between the two signals reaching me should be:
T - T(P_s - P_r)/(P_s + P_r) =
T*[1 - (P_s - P_r)/(P_s + P_r)] =
T*[(P_s + P_r) - (P_s - P_r)]/(P_s + P_r) =
2TP_r/(P_s + P_r)

...and we know the time between the two signals reaching me was itself defined to be P_r, so this implies:

P_r = 2TP_r/(P_s + P_r)
P_s + P_r = 2T
T = (P_s + P_r)/2

So, that should be the actual time between signals 1 and 2 bouncing off the reflective surface. Likewise, the time between signals 2 and 3 bouncing off should be T' = (P_s + P'_r)/2. So, if we consider an event A at the midpoint of the time interval between signal 1 bouncing and signal 2 bouncing, and likewise consider an event B at the midpoint of the time interval between signal 2 bouncing and signal 3 bouncing, the time between A and B should be:

(P_s + P_r)/4 + (P_s + P'_r)/4 =
(2P_s + P_r + P'_r)/4

If we again approximate by assuming the velocity at the midpoint between two bounces is the same as the average velocity between those bounces, then for the two midpoints A and B, (change in velocity)/(change in time) would be given by:

$$4c( \frac{P_s - P'_r}{P_s + P'_r} - \frac{P_s - P_r}{P_s + P_r}) / (2P_s + P_r + P'_r)$$

Now let's see if this altered formula works. Suppose we have a rocket accelerating in the +x direction of my inertial frame, with constant coordinate acceleration of 0.5 light-years/year² (so it will have to do this for less than 2 years or it will reach light speed, but I'll pick my numbers so that the signals bounce off it before this), with x(t) = 0.25t² (in units of c=1). Suppose I am at rest at position x=1 light years, and I send out 3 light signals in the direction of the rocket, the first at t=0, the second at t=0.0001, the third at t=0.0002. So then x(t) for the first signal is x(t) = 1 - t, for the second it's x(t) = 1 - (t - 0.0001) = 1.0001 - t, for the third it's x(t) = 1.0002 - t. So, figuring out when each signal will bounce off the rocket:

First signal
0.25t² = 1 - t
t² = 4 - 4t
t² + 4t + 4 = 4 + 4
(t + 2)² = 8
t + 2 = 2.82842712474619
t = 0.82842712474619

Second signal
0.25t² = 1.0001 - t
t² = 4.0004 - 4t
t² + 4t + 4 = 4.0004 + 4
(t + 2)² = 8.0004
t + 2 = 2.82849783454045
t = 0.82849783454045

Third signal
0.25t² = 1.0002 - t
t² = 4.0008 - 4t
t² + 4t + 4 = 4.0008 + 4
(t + 2)² = 8.0008
t + 2 = 2.82856854256707
t = 0.82856854256707

So we have P_s = 0.0001, T = 0.82849783454045 - 0.82842712474619 = 0.00007070979426, and T' = 0.82856854256707 - 0.82849783454045 = 0.00007070802662. I also showed above that T = (P_s + P_r)/2 which can be rearranged as P_r = 2T - P_s, and similarly P'_r = 2T' - P₂, so this gives P_r = 0.00004141958852 and P'_r = 0.00004141605324.

So, P_s - P'_r = 0.0001 - 0.00004141605324 = 0.00005858394676
P_s + P'_r = 0.0001 + 0.00004141605324 = 0.00014141605324
P_s - P_r = 0.0001 - 0.00004141958852 = 0.00005858041148
P_s + P_r = 0.0001 + 0.00004141958852 = 0.00014141958852

So $$\frac{P_s - P'_r}{P_s + P'_r} - \frac{P_s - P_r}{P_s + P_r}$$ = 0.414266594334775 - 0.414231239767151 = 0.000035354567624

And 2P_s + P_r + P'_r = 0.00028283564176, so the revised formula gives 4*0.000035354567624/0.00028283564176 = 0.500001589672353, which is a very accurate approximation for the true coordinate acceleration of 0.5!

 

[Passionflower]

That is fine JesseM and that formula works for a constant coordinate acceleration. But how do you propose to get from your construction to the principal question?

For a particular point in time one can calculate the proper acceleration but in case of a constant proper acceleration the coordinate acceleration is decreasing.

 

[JesseM]

That is fine JesseM and that formula works for a constant coordinate acceleration. But how do you propose to get from your construction to the principal question?

For a particular point in time one can calculate the proper acceleration but in case of a constant proper acceleration the coordinate acceleration is decreasing.

In your OP you didn't say you wanted only proper acceleration, you said "how do we express the acceleration (either proper or coordinate) in terms of an increase of the rate?" My derivation doesn't make any assumption of constant coordinate acceleration, it will work fine for varying coordinate acceleration too. And as you can see from this textbook, the relation between proper acceleration a and coordinate acceleration dv/dt, $$a = \frac{1}{(1 - v^2/c^2)^{3/2}} dv/dt$$ holds generally, it doesn't depend on the assumption of constant proper acceleration. Since I was deriving the acceleration by assuming a velocity of $$c \frac{P_s - P_r}{P_s + P_r}$$ at the first midpoint A and a velocity of $$c \frac{P_s - P'_r}{P_s + P'_r}$$ at the second midpoint B, I think it should work to plug in an average velocity of $$c/2 ( \frac{P_s - P_r}{P_s + P_r} + \frac{P_s - P'_r}{P_s + P'_r})$$ into the formula, so the whole formula for proper acceleration would be:

$$\frac{4c}{(1 - (1/4)( \frac{P_s - P_r}{P_s + P_r} + \frac{P_s - P'_r}{P_s + P'_r})^2 )^{3/2}}* ( \frac{P_s - P'_r}{P_s + P'_r} - \frac{P_s - P_r}{P_s + P_r}) / (2P_s + P_r + P'_r)$$

Of course this would only give an approximate value, but in the limit as the time period between signals approached zero I'd expect this to approach the exact value of the object's proper acceleration at the moment the 2nd signal hit the object (if you want to do a numerical example, pick a time between signals that is very small compared to the magnitude of the acceleration in units where c=1, as I did in mine where the coordinate acceleration was 0.5 light-years/year² and the time between sending successive signals was 0.0001 years)