Zero relative speed of light?

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).

DaleSpam

I agree.

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.

No, it is moving tangentially relative to you.

No, see above.

Doc Al

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?)

penomade

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!
I am thinking about it.

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.

DaleSpam

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 [itex]r(t)=\sqrt{v^2 t^2 + {r_0}^2}[/itex] where t is the time before or after the point of closest approach.

[tex]r'(t)=\frac{v^2 t}{\sqrt{v^2 t^2 + {r_0}^2}}[/tex]
[tex]r''(t)=\frac{v^2 {r_0}^2}{\left(v^2 t^2 + {r_0}^2\right)^{3/2}}[/tex]

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.

penomade

Thank you Dale, I just give it a thought and return.

bahamagreen

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.

russ_watters

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