Speed of the Earth through the Universe

I heartily recommend starting with Stephen Hawking’s book, A Brief History of Time.

If you calculate the movements of all of the galaxies we can see, you’ll find that they’re (mostly) receding from us, and from each other. You could, indeed, trace them back and find an intersection point (although there are anomalies). But who’s to say that intersection point isn’t moving?

CuriousCanuck seems to be trying to move his reference point farther and farther away from us to get a more stable answer. The problem is that our current observable data shows that the farther you get from Earth, the faster objects are moving away from us, and they’re all traveling in different directions!

Does not work.

Remember that the Big Bang did not start as a singularity (point) floating about somewhere in space. There was no space at all. The Big Bang created the very space into which everything formed. Space itself expanded…not just an explosion flinging bits into space which was already there (which is what you would need for your idea to work).

That needs to be repeated!

Just to emphasize what has already been said, the Big Bang happened everywhere in the universe at the same time. The “fixed point” of the Big Bang was the universe, in its entirety.

This thread makes me ask something that has been bothering me for a while…

What if you take off in a rocketship from earth, and accelerate to 1/2 lightspeed. Then you fire two lasers, one in the direction of motion, and the other opposite. Would the front beam seem (from the rocket) to travel at 1/2 the speed of light, and the one behind at 1.5 the speed of light?

If that’s the case, we could just fire lasers from all directions on Earth and see which “direction” and velocity we are travelling in respect to… uh… light’s static speed “frame” or something… factoring in the motion of the galaxy, Earth’s rotation, orbit, etc.

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Yes, it’s a very nice reference, but it’s still not “stationary” with no relative clause. It’s “stationary relative to itself”.

Ahem

NO! NO! NO! NO! NO! A THOUSAND TIMES NO!

This is the canonical example that goes right to the heart of SR. There’s not a “static speed frame”. Light’s speed is constant, and everyone sees it the same, even if they’re moving with respect to each other. If you did this experiment you suggest, all the lasers would go out at exactly the same speed.

Sure, and if it gets too much discussion, it’ll feed the misconception that there’s a ‘center’ from which the big bang sprang. It’s just a very nice reference.

I think it was Sagan, on some PBS program many yearsago, who said “Wherever you are in space that point isn’t moving.” I shouldn’t really put that in quotes, because it probably isn’t exactly right. I never really understood his point, but I think he said that even if you’re in a rocket ship, point x stays where you are, and all else moves relative to that. Something to do with the meaning of “moving.”
Maybe it was Mr. Rogers. :wink:

This is going to hurt your head…

If you measured (from the rocket) those two lasers, both of them would be traveling at exactly c (the speed of light). If you measured them from the Earth, they would still both be traveling at exactly c.

Now the real twist: Let’s say, instead of lasers, that you had two phenomenal little guns that could fire measuring devices at .99c (we can’t actually shoot them at c, for hopefully obvious reasons). You accelerate your rocketship to .5c, and you use your guns to simultaneously shoot out two measuring devices–one going “forward” (in the direction of the motion of the rocket) at .99c and the other going the opposite direction at .99c.

If you measure the speed of the two measuring devices from the rocket, they are both traveling at just under c. If you measure the speed on one of them from the other one, it will be traveling at just under c, not at just under 2c.

OK, wouldn’t be true however that they are moving apart at just under 2C, although the measurement could only be calculated at just under (or at) C?

It’s that because time slows down the closer you get to light speed?

Whassis SR it’s getting to the heart of? Special Relativity?

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SR is indeed Special Relativity. Time dilation is a consequence of the assumption that the speed of light is constant. It should be better stated: “As a clock approaches the speed of light relative to an observer, the clock will appear to slow down as measured by that observer.”

Yeeeeah, but only relative to the observer? I thought time actually slowed for the traveler… which allows the light beam he shines in front of him to to go the same velocity relative to himself.

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No. An observer moving with the clock would see it ticking along at the same pace as ever, since relative to him the clock is stationary.

You mean that they’re measured as moving apart at just under c but they’re “really” going faster.

I’ll let you think about that for a while to give you the chance to retract it on your own.

Retract what? Two objects moving apart relative to eachother at a combined speed over C does not violate the rules of physics as long as each one is not accelerating itself faster than c. Being able to measure the speed from one or the other at 2c does. There is a difference there.

Okay…

To say “each one is not accelerating itself faster than c” implies that you’re assuming the presence of a canonical frame in which to measure velocity. The combined speed will never exceed that of light because (and how many times has this been said in this thread alone?) you don’t combine speeds by adding.

Hey, I am aware of that. The formula for the above scenario would be v = (w - u)/(1 - wu/c2)
which I am aware is not adding. However we were talking about a question of light speed from the perpective of the rocket (or earth, or whatever) as a frame of reference and the differences between that and using one of the accellerating objects as a frame of reference.

From the centre stationary originating source frame of reference, the two objects flying apart from eachother could be said to be obsereved as retreating at greater than c without violating the rules of physics. Do the same measurement from one of the flying objects, and v = (w - u)/(1 - wu/c2) is used.

I mean retreating from eachother of course.