# SR question

1. Nov 28, 2013

### choran

First of all, Happy Thanksgiving to you all! Now, my question:

I'm on earth viewing through a telescope. I observe a car in outer space, traveling roughly at right angles to me (i.e., across my scope's field of view, let's say left to right). That car, a 1956 Chevy (irrelevant, but my favorite car) according to its speedometer, is traveling 1/2c. It turns on its headlights. He measures the speed of light emanating from his headlights as traveling c. I measure the light traveling from his headlights as c.

Question: What do I measure the car's speed as being?

Thanks.

2. Nov 28, 2013

### TumblingDice

Happy Turkey Day to you! The question requires more information. When you provide a rate of speed you need to also mention, "relative to XXX.". I mention this just to be clear about the 'speedometer' you used to provide the measurement.

For this thought experiment, if the driver is measuring his speed relative to Earth, then you measure his speed at .5 c also. But you need to be clear that everything is relative. The speedometer is meaningless without any 'road' that it's measuring against. There is no 'absolute' speed or any preferred reference frame in relativity. It's important to gain that understanding before moving forward or nothing will make sense.

3. Nov 28, 2013

### Psychosmurf

That depends on the velocity of whatever the car is driving on. In that case you use the velocity-addition formula. If, however, the car is out in empty space, then its speedometer would have to measure its speed as 0, not 1/2 c.

4. Nov 28, 2013

### choran

OK, gotcha. Let's say it's driving on a track suspended in space. The car's speed of 1/2c is relative to that track.
To us on earth, can we say that the track appears motionless? Thanks!

5. Nov 28, 2013

### choran

Re: Psycosmurt.
Can we posit that the track upon which the car is driving is motionless with respect to earth, and the car moves .5 c relative to that track, and thus relative to earth?

6. Nov 28, 2013

### TumblingDice

Motionless to you would mean the road is in your inertial reference frame. The speedometer reads .5 c relative to the road (and you). You measure the car and driver's speed at .5 c also, relative to you (and the road).

7. Nov 28, 2013

### choran

Tumbling Dice! One of my favorite song! LOL
Well, I think you just made my buddy very happy. I was convinced that the earthbound guy would measure the car's speed as other than .5 c, for some reason. You are right, though--we seem to be in the same reference frame.
OK, then here's my question, now that my buddy's happy: Under what conditions, keeping the facts as close to the above as possible, would I measure the car as OTHER than .5 c?
Thanks

8. Nov 28, 2013

### Staff: Mentor

We can. And in that case you on the earth measure the car's speed to be .5c relative to you, the earth, and the track, of course.

The speed of the light from the headlights of the car is c relative to the car and relative to you. This result is consistent with the relativistic velocity addition formula $w=\frac{u+v}{1+uv}$, where $u$ is the speed of the car relative to you, $v$ is the speed of something relative to car, and $w$ is the speed of that something relative to you.

9. Nov 28, 2013

### TumblingDice

Well, one way would be to all out nit-pick and bring up gravitational fields, the rotation of the Earth, or other hard core calculations that neither of you considered as premises when you created the scenario.

If I understand the spirit of the question, the only way for you to measure different than .5c would be if the track was in constant motion relative to you, to keep it simple (so it remains an inertial reference frame w/o accelerating relative to you). In that case the addition of velocities calc would be used for the speedometer and track to obtain relative to you. The speeds DO NOT just add or subtract as intuition might lead you.

Here could be a way you could concede to the first scenario, but get a chance to regain poise with a second question. Say the track is moving at .25c relative to you, and the speedometer reads .5c. How fast do you measure the car as moving? Now that's a good question to learn more!

10. Nov 28, 2013

### choran

Yes, that's the question: If the track is in motion relative to the observer on earth at .25c, and the car is in relative motion to the track at .5 c, then the observer on earth would measure the speed of the headlights at c, but measure the speed of the car as: ??
I assume I'd have to apply a time dilation formula of 1/square root of 1-v2/c2. I'm too old. Not worried about poise. LOL Do the math for me and I'll buy you a shot of Seagrams 7! LOL

11. Nov 28, 2013

### TumblingDice

Nugatory posted the calc above when mentioning the speed of the headlamp beams. The value for c in this case is set to '1', so...
(.5 + .25) /( (1 + (.25 x .5)) =
.75 / 1.125 =

.666666 c !

If there's a math error, I get points for showing my work.

12. Nov 28, 2013

### choran

OK, who do I send the Seagrams to? Thanks fellas, appreciate it! My buddy was right! OK, here's my last one, and it's for me. Don't yell, now! Remember, I'm old and not about to return to college--been there, done that, and had a fine career in another profession, now retired having a ball (you'll get there someday). BUT, I'm worried (can't enjoy my Seagrams tonight) because of Herbert Dingle (don't cringe!) and his idea that under SR, if A sees B's clock as slow, B will also see A's clock as slow. Talking straight SR now, no preferred frames. How can both be right, and what's the tiebreaker? When it's time for the showdown "Gentlemen, produce thy watches", whose watch is slow? Thanks for the help, and I admire your diligence in pursuing this most difficult of fields.
Thanks
CH

13. Nov 28, 2013

### Staff: Mentor

If both are inertial then there is no tiebreaker. B's clock runs slow in A's frame and A's clock runs slow in B's frame.

14. Nov 28, 2013

### Psychosmurf

Ask this question: Why can't they both be right? They are running two totally different experiments, and there is no reason at all a priori to think their measurements should be exactly the same. In fact, if spacetime had no geometry, their time measurements would be random (and meaningless), though there would be no reason to believe that one is any more right than the other. However, spacetime does have a geometry, but it turns out that this geometry is Minkowskian (that is, the time measurements of two inertial observers will be related by the time-dilation equations, instead of just being random) rather than Euclidean (as we like to assume). The belief that they should both be the same or that one is somehow more right than the other comes from our Euclidean prejudices and nothing more.

15. Nov 28, 2013

### choran

When you say "if both are inertial", do you mean assuming both are noting in uniform relative motion relative to one another? I guess that's the answer, but a tough one to swallow! Like saying Joe is fatter than Al, but Al is fatter than Joe. Half a shot of Seargrams for Dale! LOL

16. Nov 29, 2013

### Staff: Mentor

Try this post: https://www.physicsforums.com/showpost.php?p=4556116&postcount=16

They're both slow and there's no paradox.

17. Nov 29, 2013

### dvf

No, more like, from a distance you look smaller to me, and I look smaller to you.

18. Nov 30, 2013

### Staff: Mentor

Yes.

Well, you have to be careful in expressing things in relativity. Usually, when there is some "both are fatter" statement it is because you have made a frame-variant statement without specifying the frame. If you specify the frame of every frame-variant quantity, then it becomes clear that you are not making a self-contradictory statement.

19. Nov 30, 2013

### choran

Well, still hard for me to internalize that one. Certain can see exactly what you are saying, it sounds good. Can't get over that at the exact moment A is measuring B's clock as slow, B at that exact moment is measuring A's clock as slow. It's not a matter of perspective--that would be easy to accept, as in the height analogy dvf gave above.
In the clock situation, we can't say "Freeze and let me back off and I'll tell you who's taller, as in the height analogy." Can't say "freeze, show your watches" and declare a winner. Not at all hard for me to see how in various real world scenarios A sees B's as slower. What is hard is the idea of symmetry, if that's the right word.

The concept (still talking SR only) that if A and B are moving relative to one another, A has just as much right to consider himself as stationary as B does, and vice versa, leading to what I just can't swallow. Rocket leaves earth--universe did not get up and move away from rocket. Earth did not pick up and go--the rocket left.
Muon example--Rocky Mountains did not rush up to meet the muon, no matter how much a conceited muon might
believe that. lol Rocky mountains did not become foreshortened, etc. Muon decayed at speed. Muon might measure my clock as slow, but the muon, I believe, would be wrong. Oh, well, back to the same old problem.
I think one has to go through physics boot camp, have a D.I. for several years, and then this stuff is internalized.
Otherwise, no shot. Thanks again, buddy.

20. Nov 30, 2013

### Staff: Mentor

Perhaps this will help a little. For a moment, let's consider only Newtonian physics, i.e. Galileo's relativity, not Einstein's.

Now, in Newtonian physics it makes sense to say that Al is taller than Joe without any qualifiers, this is because length is frame invariant. It doesn't matter if you do the comparison in Al's frame, Joe's frame, or the earth's frame, the answer is the same.

However, in Newtonian physics it does not make sense to say that Al has more speed than Joe without any qualifiers, this is because speed is frame variant. In Al's frame Joe has more speed, and in Joe's frame Al has more speed, and perhaps in the earth's frame they have the same speed.

So the statement "Al has more speed than Joe" is fundamentally meaningless because we have stated a frame-variant quantity, speed, without specifying the frame it refers to. In contrast, the statement "Al has more speed than Joe in Joe's frame" is meaningful because we have specified the reference frame refered to by the frame-variant quantity. Also, the fact that Al has more speed than Joe (in Joe's frame) does not in any way contradict the fact that Joe has more speed than Al (in Al's frame). Speed is a frame variant quanitity and the two different statements are made wrt two different frames.

You can clearly see that even in Galileo's relativity there are frame variant and frame invariant quantities. You can also see that even in Galileo's relativity you have to specify the frame for frame variant quantities and you can get seeming contradictions if you drop the frame specification.

The difference between Einstein's and Galileo's relativity is not how invariant and variant quantites are treated. It is merely that the list of frame-variant quantities is larger in Einstein's relativity. So some things, such as length, duration, and simultaneity, which you could previously get away without specifying the frame, you now need to specify the frame. It isn't a big conceptual change, just a shuffling of some items from the "frame invariant" list to the "frame variant" list.