1.5 Times the Speed of Light

Doc Al

Right!

Nothing absolute about the ground frame, that's for sure!

What do you mean by 'forget the ground'? You gave distances and times that only make sense in the frame of the ground! Specifying a distance without mentioning the frame you are using is meaningless, right? (Unless you secretly do believe in some absolute frame!)

Again, that's the closing speed according to the ground frame. You can't 'forget the ground'.

Bussani

You can't say, "Forget the ground," and then use distances according to the ground.

DrGreg

Speed relative to a frame is measured by distance in the same frame divided by time in the same frame.

You are mixing frames.

James S Saint

Yes, but I'm not talking about what a ground observer sees. I see the distance of 2Ls get reduced to 0 in only 1.333 secs. That is what matters. Perhaps I cannot actually see the ground at all or even know it exists. As stated just prior, the ground only serves as an equal frame for us to begin and end. Both of us would end up seeing that 2Ls of distance vanished in only 1.333 secs.

All observers would see us meet at the sign. The ground person would see each of us traveling .75c toward the sign. We would each see ourselves approaching the sign at .75c. But regardless of what we might observe of each other, our end measurements would be that we reached each other in only 1.333 secs causing a 2Ls distance to be traversed.

espen180

You are - whether you ralize it or not - working in the ground frame, in which the relative speed of the ships is only constricted to be smaller than 2c, and in which your result is correct. You cannot, however, extrapolate the measurement of the ground frame to the ship frame without using the Lorentz transformation equations, which will tell you that according to an observer on either of the ships, the other ship is approaching at .96c.

HallsofIvy

No, it doesn't. If you and he were traveling in the same direction, at the same .7t c, then it would be
[tex]\frac{.75c- .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= 0[/tex], of course.

James S Saint

No it is according to US.

We measure 2Ls of distance. In merely 1.333 secs, we measure NO distance between us. How is that not 1.5c?

jcsd

But think what happens in the ground observers frame when one of you shines a light at the other, the light, travelling at the speed of light, reaches the other before the pair of you meet.

Is the light therefore travelling at greater than 1.5 times the speed of light? That would be totally illogical as we started by assuming it travels at the speed of light.

HallsofIvy

Wow, 57 posts in less than 3 hours! That may be a record. James S Saint, you are doing all of your computations as if Newtonian physics applied. Naturally, you are going to get a contradiction to relativity.

James S Saint

To put it simply (I hope)

We measure 2Ls of distance.

1.333 secs later, we measure 0 distance.

How is that possible?

Doc Al

Who is "US"? You? Your brother? The ground? All three frames will measure different distances and times.

You say "US", but you are actually using ground frame measurements.
It is 1.5c, but only in the ground frame. You used distances and times in the ground frame, so the closing speed is in the ground frame.

Austin0

They would all agree on the simultaneity of the intersection. If they don't steer carefully they could headon and then it would be obvious they met simultaneously. Seriously though, all observers agree on local events at a single location. SImultaneity becomes relevant when events are separated spatially

Doc Al

Where's the problem?

You are confusing a closing speed greater than c with something actually moving with a speed greater than c in some reference frame. But that's not the case.

James S Saint

Actually I'm tickled to find a forum where people are so active.

Now if I can just get them to follow logic... grin. :)

jcsd

Becaus we're measuring the distance and time in the ground observers frame, where both of you are moving. Simples.

Imagine a 100m race. If someone runs 100m in 10 secs (assuming constant speed), but the finsihing linen approaches them at the same speed, do we say they ran 100 m in 5 secs? No we don't bceause we're talking about the frame in which both the runner and the finsihing line are moving,therefore the runner ran 50m in 5 secs.

Borg

Also, what about someone standing at the one second mark and in the signpost's reference frame. If your brother shines a light back at that observer after he passes, will that observer see light travelling at 1/4 the speed of light? No.

Doc Al

You abuse the term 'logic'. What you really are saying: "If I can just get them to agree with me." But many of your statements are incorrect!

It's not a problem with 'logic', but with your understanding of physics.