Register to reply 
Distance between fast moving objects 
Share this thread: 
#1
Sep612, 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. 


#2
Sep612, 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. 


#3
Sep612, 10:58 AM

P: 52




#4
Sep612, 05:31 PM

Sci Advisor
PF Gold
P: 1,848

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 thereandback 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] 


#5
Sep612, 07:29 PM

P: 52




#6
Sep612, 08:42 PM

PF Gold
P: 4,776

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.



#7
Sep612, 09:27 PM

P: 52




#8
Sep612, 10:04 PM

PF Gold
P: 4,776

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. 


#9
Sep712, 05:22 AM

P: 429

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.



#10
Sep712, 03:27 PM

#11
Sep712, 04:58 PM

Emeritus
Sci Advisor
P: 7,657




#12
Sep712, 05:45 PM

P: 30

Velocities according to an accelerating observer are discussed in this website: https://sites.google.com/site/cadoeq...eferenceframe 


#13
Sep812, 11:41 AM

PF Gold
P: 4,776

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. 


#14
Sep812, 01:26 PM

P: 30




#15
Sep1212, 04:46 PM

#16
Sep1312, 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?



#17
Sep1312, 04:37 PM

PF Gold
P: 4,776

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. 


#18
Sep1312, 06:31 PM

P: 52




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 