# Zero relative speed of light?

1. Jan 7, 2013

Suppose a beam of light is approaching you from a distant source. As it comes closer to you, it doesn’t hit you; instead it misses you by a 10 meter distance. How can the rate of acceleration of that beam relative to you, be explained(in non-jargon language)? What would the speed of that be, when it reaches the closest distance of 10 meters, relative to you?

2. Jan 7, 2013

### A.T.

What acceleration? Light moves at a constant speed.
299792458 m/s

3. Jan 7, 2013

This is on the presumption that because that beam of light travels at an angle, its speed reduces as it gets closer to you. In fact this is my question.

4. Jan 7, 2013

### A.T.

No, it doesn't.

5. Jan 7, 2013

### DiracPool

You can't make a presumption like that, unless you're writing a sci-fi novel. Light travels in straight lines only and at one speed only (in free space). This rule neglects general relativity effects, of course, but I don't think that's what you are referring to or you would have mentioned that.

6. Jan 7, 2013

At the point it reaches the 10 meter distance, it doesn't get any closer to you. That means it doesn't move relative to you. And in a minute fraction of time later, it would be traveling away from you.

Last edited: Jan 7, 2013
7. Jan 7, 2013

### WannabeNewton

It isn't instantaneously at rest in your frame; it is moving uniformly in your frame. Light has no defined inertial frame so you can't perform a boost to some frame such that you are co - moving with light.

8. Jan 7, 2013

I don't mean the beam is behaving strangely. I just mean instead of coming straight on to you, it passes by you. this is what I mean by an angle.

9. Jan 7, 2013

### Staff: Mentor

The speed of the beam is always c and the acceleration is 0. However, I think that what you are asking is what are the first and second time derivatives of the distance to the beam. You can get that simply by transforming to polar coordinates and calculating dr/dt etc. Those quantities are not physically meaningful, merely a reflection of the coordinates, i.e. there is no real force causing an acceleration.

10. Jan 7, 2013

Perhaps I,d better put it this way, at first the beam is coming on to you. at that period it has a positive speed. then when it's leaving you, it has a negative speed, as it were. In between there should be a moment that it's neither getting closer nor getting away from you(relative to you). what's that?

11. Jan 7, 2013

Let's forget about acceleration, for the moment. Instead, I use the word speed. what about that? Is it true that at one moment a photon is stationary relative to me.

Last edited: Jan 7, 2013
12. Jan 7, 2013

### Staff: Mentor

No, it is not true that at any moment a light pulse is stationary relative to you. What is true is that at some moment the light pulse is moving tangentially to you. It is also true that at the same moment the change in the light pulse's distance to you is 0.

13. Jan 7, 2013

### DiracPool

As it IS, there is no such thing as negative speed. There is negative velocity in relation to position.

THAT, is exactly what that is. It's neither getting closer nor farther from you at that specific instant it is passing you. It's hard to see exactly what your quandry is, but I think you don't have a grasp on the special nature of what light is, it doesn't behave in the same manner a car does when it whizzes by you, which I think you're trying to equate it to.

14. Jan 7, 2013

### Staff: Mentor

I don't think that his confusion is related specifically to light. I think that it is more of a confusion about definitions of speed and velocity. If you are standing on a sidewalk next to a highway then the cars are travelling at ~100 km/h velocity relative to you. At no time is their velocity 0, and at no time is their speed 0. However, the distance to them is first decreasing, then momentarily constant, then increasing.

15. Jan 7, 2013

### DiracPool

I think I might be getting a better fix on the question now. The problem with the question is why you are transfixed on when a photon is stationary relative to you. What deeper meaning to this are you looking for?

The answer is that it is true all the time. Any time you want to take an instantaneous snapshot of space time, yes, there it is, the photon is stationary relative to you. It doesn't matter if it is 100 light years, 100 miles or 10 meters away. When it is approaching you, it approaches you at the speed of c with a positive velocity, if you will, and when it leaves you, it also leaves at c, relative to you. The only thing that changes that instant it passes you is the position value, which would be zero (forget about the ten meters, just have it pass right through you). The moment it passes you, though, "zero dark photon," has no effect on intertial reference frames or relative motion so I'm not sure where you are going with it.

16. Jan 7, 2013

### WannabeNewton

If you define in your inertial frame the +x direction to be the direction the massive particle is traveling in at constant speed then v = ||v||i where i is the usual cartesian unit vector. The speed ||v|| is constant so nowhere does it suddenly become zero unless it is zero at t = 0 when you first measure its speed. You don't even need to look at light.

17. Jan 7, 2013

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.

18. Jan 7, 2013

Maybe you can help me find a valid answer. I will also present my humble solution.(which I have partly done).

Last edited: Jan 7, 2013
19. Jan 7, 2013

### Staff: Mentor

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.

20. Jan 7, 2013

### Staff: Mentor

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?)

21. Jan 7, 2013

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!

22. Jan 7, 2013

### Staff: Mentor

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.

23. Jan 7, 2013

### Staff: Mentor

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.

Last edited: Jan 7, 2013
24. Jan 7, 2013