# Zero relative speed of light?

Tags: light, relative, speed
Mentor
P: 17,322
 Quote by penomade If you agree that speed involves movement
I agree.

 Quote by penomade , and movement involves distance,
I disagree. Movement involves a change in position wrt time, not a change in distance wrt time. This is the source of your confusion, I believe.

 Quote by penomade Although the movement of the car is tangential, but at the very moment it is, on its path, at the perpendicular line of my sight, it does not have the slightest movement relative to me.
No, it is moving tangentially relative to you.

 Quote by penomade That's, after all, what the meaning of movement is: "changing distance". therefore, it is, just for one moment, stationary relative to me.
No, see above.
Mentor
P: 41,475
 Quote by penomade Granted. Let's talk about your example first. If you agree that speed involves movement, and movement involves distance, then if there is no change of distance, than there is no movement. Although the movement of the car is tangential, but at the very moment it is, on its path, at the perpendicular line of my sight, it does not have the slightest movement relative to me. That's, after all, what the meaning of movement is: "changing distance". therefore, it is, just for one moment, stationary relative to me.
Just because, for an instant, it is not getting closer to you does not mean it is stationary relative to you. If you are measuring the light's position using the radius vector with respect to you, realize that there are two components to the velocity: the radial component (which would be zero at that moment) and the tangential component (which equals c, as usual).

(If you really believe that the speeding car does not have the slightest movement relative to you, why not reach out and touch it?)
P: 14
 Quote by Doc Al (If you really believe that the speeding car does not have the slightest movement relative to you, why not reach out and touch it?)
Interesting question. I think you put it politely. May be what you mean is, better said, why don't I go and stand exactly at the foot of the perpendicular!
Mentor
P: 17,322
 Quote by penomade What made me think of this, is this alleged paradox.http://www.phys.unsw.edu.au/einstein...le_paradox.htm Maybe you can help me find a valid answer. I will also present my humble solution.(which I have partly done).
The resolution to this "paradox" is the relativity of simultaneity (as it usually is). The barn doors are closed at the same moment in the barn's frame, but in the right door closes much earlier than the left door in the pole's frame.
 Mentor P: 17,322 Just for reference, if an object is moving in a straight line at speed v wrt me and its closest point of approach is a distance r0 from me then the distance from me is $r(t)=\sqrt{v^2 t^2 + {r_0}^2}$ where t is the time before or after the point of closest approach. $$r'(t)=\frac{v^2 t}{\sqrt{v^2 t^2 + {r_0}^2}}$$ $$r''(t)=\frac{v^2 {r_0}^2}{\left(v^2 t^2 + {r_0}^2\right)^{3/2}}$$ Again, r' is the time rate of change of the distance between the object and me, it is not the speed of the object wrt me. That is v.
P: 14
 Quote by DaleSpam The resolution to this "paradox" is the relativity of simultaneity (as it usually is). The barn doors are closed at the same moment in the barn's frame, but in the right door closes much earlier than the left door in the pole's frame.
Thank you Dale, I just give it a thought and return.
 P: 543 The OP's question is a good one and clearly shows an intuitive reach that exceeds his grasp of some basics... just needs clarification of fundamentals about lengths and times - values, deltas, infinitesimal differentials, and instantaneous values.
Mentor
P: 22,300
 Quote by DaleSpam Again, r' is the time rate of change of the distance between the object and me, it is not the speed of the object wrt me. That is v.
Let me try to back up a bit and talk about coordinates/reference frames:

A reference frame is a set of xy (and maybe z) coordinates. Speed is the change in position in that coordinate system (divided by time). If you happen to be located at the origin of that coordinate system, the object you are observing does not have to be moving directly toward or away from you to have a measurable speed. Consider, for example, constant speed motion from (3,0) to (0,3) in a coordinate system.

Examples of 1 dimensional motion are often used to simplify problems, but that shouldn't trick one into thinking motion is all directly toward or away from the origin of the reference frame.

In the Navy, we used what is called a "maneuvering board" to transform bearing and distance to a target (polar coordinates) to two different Cartesian coordinate systems centered on our ship, to calculate the distance and speed of another ship relative to us (and closest point of approach), in our moving frame of reference and in a frame stationary with respect to the water: http://gcaptain.com/maritime/tools/files/MoBoard.pdf

 Related Discussions Special & General Relativity 26 Special & General Relativity 17 Introductory Physics Homework 2 General Discussion 16 General Physics 9