Register to reply

Distance between fast moving objects

by Chuck37
Tags: distance, moving, objects
Share this thread:
Chuck37
#1
Sep6-12, 10:43 AM
P: 52
The equations for length contraction are clear when it comes to something like a ruler flying past, or equivalently, two objects flying in formation. I'm not sure how to think about distance to a moving object, or between two moving objects. Is there an easy way to think about this?

Two examples for discussion purposes: (1) I'm stationary and an object is moving rapidly away. What does relativity do in terms of my perception of his distance vs. time? Maybe I'm using laser pulses to measure range. (2) I'm in a third frame looking at two objects moving apart at high speed. How do I perceive their separation?

Thanks for any pointers.
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
Muphrid
#2
Sep6-12, 10:48 AM
P: 834
Length contraction only really happens because you have some extended object and you're looking at the distance between both ends of the object. So you can't really see anything of the sort with the distance to such an object.

Similarly, if two objects are moving apart, it's not immediately clear to me that you could talk about contraction of that distance, either, because they don't constitute a single extended object. It's possible to analyze this case mathematically; it just doesn't fit the simple case of an extended object that's moving and would obviously length contract.
Chuck37
#3
Sep6-12, 10:58 AM
P: 52
Quote Quote by Muphrid View Post
Length contraction only really happens because you have some extended object and you're looking at the distance between both ends of the object. So you can't really see anything of the sort with the distance to such an object.

Similarly, if two objects are moving apart, it's not immediately clear to me that you could talk about contraction of that distance, either, because they don't constitute a single extended object. It's possible to analyze this case mathematically; it just doesn't fit the simple case of an extended object that's moving and would obviously length contract.
Thanks for the reply. There must be something here, but maybe it's too complicated to just apply a simple equation. Try this thought experiment. Two objects are flying in formation at 0.5c. Surely I can apply the length contraction equation to find the difference in how each of them perceives their separation and how someone in a stationary frame perceives it, right? Now change one guy to 0.49999c. The answer will be close, but not exactly the same. how would I compute the new answer?

DrGreg
#4
Sep6-12, 05:31 PM
Sci Advisor
PF Gold
DrGreg's Avatar
P: 1,843
Distance between fast moving objects

First, from a practical point of view, you can measure distance from yourself (an inertial observer) using radar. The total there-and-back time gives you twice the distance (when you multiply by c), and the time at which that distance is valid is halfway between transmission and reception of the radar pulse.

Secondly, the reliable way to change coordinates from one inertial observer to another is to use the Lorentz transform[tex]
\begin{align}
t' &= \gamma (t - vx/c^2) \\
x' &= \gamma (x - vt)
\end{align}
[/tex]where[tex]
\gamma = \frac{1}{\sqrt{1 - v^2 / c^2}}
[/tex]One you've worked out distances and times in one frame you can use the above equations to convert to another frame. For example, consider[tex]
\begin{align}
x_A &= 0.5 \, c t \\
x_B &= L + 0.49999 \, c t
\end{align}
[/tex]
Chuck37
#5
Sep6-12, 07:29 PM
P: 52
Quote Quote by DrGreg View Post
First, from a practical point of view, you can measure distance from yourself (an inertial observer) using radar. The total there-and-back time gives you twice the distance (when you multiply by c), and the time at which that distance is valid is halfway between transmission and reception of the radar pulse.

Secondly, the reliable way to change coordinates from one inertial observer to another is to use the Lorentz transform[tex]
\begin{align}
t' &= \gamma (t - vx/c^2) \\
x' &= \gamma (x - vt)
\end{align}
[/tex]where[tex]
\gamma = \frac{1}{\sqrt{1 - v^2 / c^2}}
[/tex]One you've worked out distances and times in one frame you can use the above equations to convert to another frame. For example, consider[tex]
\begin{align}
x_A &= 0.5 \, c t \\
x_B &= L + 0.49999 \, c t
\end{align}
[/tex]
Thanks for the reply. The gist of my question was whether I need to consider relativity when doing the radar trick if I want to get extremely accurate answers (assuming my radar is good for it, of course). I more or less understand how to move between frames, but it's not obvious whose frame to be in, radar or object. I need to think more about your answer too.
ghwellsjr
#6
Sep6-12, 08:42 PM
PF Gold
P: 4,695
If you use radar and do the calculations that DrGreg described, then you have used the definitions of Special Relativity and selected your rest frame as the frame in which the object is moving.
Chuck37
#7
Sep6-12, 09:27 PM
P: 52
Quote Quote by ghwellsjr View Post
If you use radar and do the calculations that DrGreg described, then you have used the definitions of Special Relativity and selected your rest frame as the frame in which the object is moving.
If I measure the time for a pulse to travel round trip and multiply by c/2, in what sense am I using special relativity? That seems like the plain vanilla way to me. Maybe I missed your point.
ghwellsjr
#8
Sep6-12, 10:04 PM
PF Gold
P: 4,695
As I said, you're using the definitions of Special Relativity.

EDIT: You were wondering what frame the measurement/calculation applied to. That's what I was answering.
grav-universe
#9
Sep7-12, 05:22 AM
P: 424
SR is the same as Galilean when measuring the distances to objects from a single reference frame when constant speeds are involved. It is just d = v t. The same applies for the distances between objects. However, the speeds of objects are measured differently from different frames, found by applying the relativistic formula for speeds, but that is only necessary if transforming the measurements to a different frame of reference.
GrammawSally
#10
Sep7-12, 03:27 PM
P: 30
Here's what I got, for the two accelerating travelers problem (for one particular choice of the parameters):



The scenario involves two travelers, initially mutually stationary, and initially separated by 0.5 lightyears. Traveler 1 is initially located at the spatial origin, X = 0, of an inertial reference frame, and Traveler 2 is initially located at X = 0.5 ly in that same inertial frame. At the initial instant (T = 0), I've arbitrarily taken both of their ages to be zero years old. There is another person (the "Home-person") who is permanently located at the spatial origin, and who is also zero years old at the initial instant. So the given inertial frame represents the viewpoint of the Home-person.

At the instant T = 0, both travelers accelerate at one ly/y/y (which is just slightly less than one "g" ... 0.970g), and they both continue that acceleration throughout the scenario. The acceleration is in the direction of positive X, so Traveler 1 accelerates in the direction from Traveler 1 toward Traveler 2. And Traveler 2 accelerates in that same direction.

The plot is a Minkowsky diagram, showing the viewpoint of the Home-person. In the plot, the usual convention for Minkowski diagrams (with T plotted vertically, and X plotted horizontally) is not followed: I always prefer to plot X vertically and T horizontally.

The two curved lines in the plot are the worldlines of Traveler 1 (the lower curve) and of Traveler 2 (the upper curve). The plot shows the first two years of the Home-person's life, and it shows spatial locations out to 2 lightyears away from the origin. According to the Home-person, the separation between the two travelers is always 0.5 lightyears.

The tic-marks on the two curved lines show the ages of the two travelers. The diagram shows their lives up to (almost) 1.4 years old. According to the Home-person, the two travelers always have the same ages.

The two straight lines in the plot are the lines of simultaneity for Traveler 2, at two different choices of her age: at 0.5 years old and at 1.3 years old. These lines of simultaneity are determined using the CADO reference frame for Traveler 2. The CADO equation itself cannot be used, because the distant person of interest (Traveler 1) isn't unaccelerated. But the CADO reference frame can still be used, via either a graphical construction (as is done here), or via an iterative numerical process.

The points where these two straight lines intersect Traveler 1's worldline correspond to the age and position of Traveler 1, according to Traveler 2, when Traveler 2 is 0.5 years old and 1.3 years old, respectively.

From the diagram, we can read that, when Traveler 2 is 0.5 years old, Traveler 1 is about 0.23 years old, and is at a distance of about 0.58 ly from Traveler 2. Similarly, when Traveler 2 is about 1.3 years old, Traveler 1 is about 0.55 years old, and is at a distance of about 0.89 ly from Traveler 2.

So, for this particular scenario, Traveler 2 concludes that Traveler 1 starts out 0.5 ly away at the beginning, then is slightly farther away (0.58 ly) half a year later (in Traveler 2's time), and then is farther away still (0.89 ly) after 1.3 years. So Traveler 2 concludes that he is outrunning Traveler 1. I suspect that their separation approaches (but never exactly reaches) 1 ly, no matter how long the acceleration lasts.
pervect
#11
Sep7-12, 04:58 PM
Emeritus
Sci Advisor
P: 7,599
Quote Quote by Chuck37 View Post
If I measure the time for a pulse to travel round trip and multiply by c/2, in what sense am I using special relativity? That seems like the plain vanilla way to me. Maybe I missed your point.
I"d say this is relativistic because you've assumed (in accordance with relativity) that the speed of light is constant and equal to 'c' without regard to picking any "special" observer - we know that the speed of light is always "c", for any observer.
GrammawSally
#12
Sep7-12, 05:45 PM
P: 30
Quote Quote by pervect View Post
[...]
we know that the speed of light is always "c", for any observer.
That's true for any INERTIAL observer. It's not generally true for an accelerating observer.

Velocities according to an accelerating observer are discussed in this website:

https://sites.google.com/site/cadoeq...eference-frame
ghwellsjr
#13
Sep8-12, 11:41 AM
PF Gold
P: 4,695
Quote Quote by GrammawSally View Post
Here's what I got, for the two accelerating travelers problem (for one particular choice of the parameters):
The OP did not ask about two accelerating travelers.

The OP did not even ask about one accelerating traveler.

The OP asked about himself (or another observer) being stationary with one or more INERTIALLY moving objects.
GrammawSally
#14
Sep8-12, 01:26 PM
P: 30
Quote Quote by ghwellsjr View Post
The OP did not ask about two accelerating travelers.
You're right. Sorry ... I misread the original posting.
GrammawSally
#15
Sep12-12, 04:46 PM
P: 30
I discovered that the "tic-marks" on the two lines-of-simultaneity in the previous image weren't correct. Here's a corrected image:



With the new (corrected) tic-marks on the two lines-of-simultaneity, it can be seen that when Traveler 2 is 0.5 years old, Traveler 1 is 0.24 years old, and is at a distance of 0.53 ly from Traveler 2.

When Traveler 2 is 1.3 years old, Traveler 1 is 0.53 years old, and is at a distance of 0.67 ly from Traveler 2.

I still suspect that their separation approaches 1 ly as Traveler 2 gets older and older, but I need to investigate that further before I can confirm it.

Sorry for those tic-mark mistakes.
Chuck37
#16
Sep13-12, 02:36 PM
P: 52
Bringing this back up... if I'm moving very fast and using a radar to detect objects ahead, is there not a spatial dilation in my direction of travel that would cause me to get the wrong measurements? This is assuming my desire is to share information with someone not moving fast: "Object detected X meters ahead". Would I measure a distance √(1-(v/c)^2) short?
ghwellsjr
#17
Sep13-12, 04:37 PM
PF Gold
P: 4,695
Quote Quote by Chuck37 View Post
Bringing this back up... if I'm moving very fast and using a radar to detect objects ahead, is there not a spatial dilation in my direction of travel that would cause me to get the wrong measurements? This is assuming my desire is to share information with someone not moving fast: "Object detected X meters ahead". Would I measure a distance √(1-(v/c)^2) short?
First off, you are describing a spatial contraction, not a spatial dilation (dilation means getting bigger).

Secondly, you always have the problem when applying situations involving fast moving observers, that you just can't say "Object detected X meters ahead". What you can say is that in a particular frame of reference, you detected an object X meters ahead at some time in your past. If you have reason to believe that the object is inertial and you also used your radar to calculate its speed, then you can predict where it will be at any future time.

Thirdly, if your purpose is to communicate the location and trajectory of that object to someone else that is not at rest with respect to you, the two of you will have to have previously agreed on the reference frames you plan to use so that your information will be meaningful.

Fourthly, keep in mind that it will take time for your radio signal to reach that other person so the information he gets will be even further in the past than when you got it.

So the bottom line is that even though distances in your rest frame are contracted, you can still transform them into another rest frame in a meaningful way.
Chuck37
#18
Sep13-12, 06:31 PM
P: 52
Quote Quote by ghwellsjr View Post
First off, you are describing a spatial contraction, not a spatial dilation (dilation means getting bigger).

Secondly, you always have the problem when applying situations involving fast moving observers, that you just can't say "Object detected X meters ahead". What you can say is that in a particular frame of reference, you detected an object X meters ahead at some time in your past. If you have reason to believe that the object is inertial and you also used your radar to calculate its speed, then you can predict where it will be at any future time.

Thirdly, if your purpose is to communicate the location and trajectory of that object to someone else that is not at rest with respect to you, the two of you will have to have previously agreed on the reference frames you plan to use so that your information will be meaningful.

Fourthly, keep in mind that it will take time for your radio signal to reach that other person so the information he gets will be even further in the past than when you got it.

So the bottom line is that even though distances in your rest frame are contracted, you can still transform them into another rest frame in a meaningful way.
I think you answered, but let me be more specific. There is a radar corner reflector on the ground at a known latitude and I'm in a supersonic jet flying towards it from the east. I know where I am using a GPS and I want to tell someone on the ground where the reflector is - I just need to find longitude. When I make my light time measurement, should I account for spatial contraction in determining where the object is in terms of earth coordinates?


Register to reply

Related Discussions
Help calculating distance between two moving objects General Physics 38
Temperature of fast objects decreases?! Special & General Relativity 9
Massive objects moving too fast Special & General Relativity 57
Do we observe fast moving objects in the universe? General Physics 12
Distance problem, Using Launcher need to find objects distance Introductory Physics Homework 2