Speed of light, mass, time questions

Factual Questions
21

[QUOTE=Mosier]
I don’t understand the math, but does it say that two objects traveling at 1/2 the speed of light directly away from each other WOULD see photons emitted from each other?
[/quote]
Yes.

[QUOTE=Mosier]
I thought relativistic speeds of more than C is what defines the “observable” universe from the unobservable part, because when relative velocity exceeds the speed of light, we will never see the object at the other end.
[/QUOTE]
The observable universe is bound by the expansion of spacetime itself; once its rate of expansion (which is a function of distance from the observer, which is its own special form of total weirdness) exceeds c (or at least very, very close to it) the light beyond it is redshifted beyond visibility, and indeed theoretical detection, and it becomes, for all intents and purposes, unobservable for anyone who doesn’t have access to the sort of technology that lets Captain Kirk bang Orion slave girls from here to the Mutara Nebula on the same stardate.

Note that spacetime is not a “thing” and is therefore not subject to the limitations stated by Chronos. In fact, the boundary of the observable universe now is about 45 billion light years away, even though the universe itself is only ~13.7 Byrs old, another seemingly curious paradox which occurs due to the fact that the light emitted when these objects were only about 40-50 million light years away and was since shifted in frequency and forced to travel progressively longer distances as space stretched out like a pregnant mother’s tummy.

This stuff is wickedly freakwonderful, even to cosmologists who become giddy as schoolgirls when discussing the topic. It’s a strangeness that even science fiction authors couldn’t conjour up, and it’s all bizarrely real and continually surprising as we get more data about the universe. So don’t feel like you’ve been left out; there is so terribly much that none of us yet know about the universe that we’re all pretty much totally ignorant, or at least stuck in a back chamber of Plato’s cave.

Stranger

22

What if two objects take off from a single point in directly opposite directions, each at 1/2 c, and I am a third observer, at rest with respect to their point of origin? For me, is the difference between their velocities going to be measured as c, or 2/5 c, or something else entirely?

-FrL-

23

[QUOTE=Frylock]
What if two objects take off from a single point in directly opposite directions, each at 1/2 c, and I am a third observer, at rest with respect to their point of origin? For me, is the difference between their velocities going to be measured as c, or 2/5 c, or something else entirely?
[/QUOTE]
1/2c, by definition.

Stranger

24

[QUOTE=Stranger On A Train]
1/2c, by definition.

Stranger
[/QUOTE]

Wait, they’re going in opposite directions at 1/2 c, so shouldn’t the difference between their velocities be c/2 - (-c/2) == c?

Or else if additivity doesn’t hold in this case, then shouldn’t the difference be 2c/5 as in a previous post o this thread?

If neither, then how do I get a difference of c/2 in this situation?

-FrL-

25

[QUOTE=Frylock]
Wait, they’re going in opposite directions at 1/2 c, so shouldn’t the difference between their velocities be c/2 - (-c/2) == c?

Or else if additivity doesn’t hold in this case, then shouldn’t the difference be 2c/5 as in a previous post o this thread?

If neither, then how do I get a difference of c/2 in this situation?
[/QUOTE]
First of all, the resultant velocity difference as seen by the spacecraft of each other should be 4/5c. My error. Now that we’ve cleared that fumble, what’s up with all of these different answers people get when doing the same calculations from different reference frames?

Your question was:
[QUOTE=Frylock]
What if two objects take off from a single point in directly opposite directions, each at 1/2 c, and I am a third observer, at rest with respect to their point of origin? For me, is the difference between their velocities going to be measured as c, or 2/5 c, or something else entirely?
[/QUOTE]
With respect to your “stationary” observer, each is moving 1/2c. Or are you asking what their apparent difference in velocity is from the view of our hypothetical privlidged observer? In that case, yes, it’s c, but this is a meaningless quantity. It’s quite possible for an observer to see apparent or illusionary “non-inertial” velocities (i.e. not measured from an inertial reference system) that exceed c, but the effect is meaningless; a photon emitted by Pram A, moving at 1/2c toward Islington, directed at Pram B, moving at 1/2c to Croydon, will still be moving at c from the reference frame of Point C, which is a rather nice French-style sidewalk cafe situated right in between. In fact, the photon will appear to be moving at c in all reference frames, neglecting the disturbing effects of the aforementioned inflation of spacetime which is only significant on somewhat larger scales than Central London. From the standpoint of observers in the reference frames of Pram A and Pram B, the relative velocities with respect to each other are 0.4c even though this leaves an outside “stationary” or hypothetical privileged scratching his head at the math.

How do we fix these differences? Ah, that’s where the clock and ruler come in; for the moving observers, their clocks are running on a different scale (which is a result of having accelerated up to relativistic speed) and their rulers, as seen by the observer at Cafe C, are somewhat shorter (and shorter still between Prams A and B, although each is convinced that their ruler is accurate and the other guy is seriously in the wrong). So, all measures of space and time (which are two components of the same metric) are affected such that any given observer assumes that he’s okay and it’s the other guy who has serious problems with his lifestyle and calibrating his watch.

The upshot is, nobody actually gets to go the speed of light or faster, even though a classical intuition would suggest that you simply point two parambulators in opposite directions and send them off at >1/2c. Which really is a good thing in the whole general scheme of things even if it does slightly complicate the math.

Stranger

26

[QUOTE=Stranger On A Train]
First of all, the resultant velocity difference as seen by the spacecraft of each other should be 4/5c. My error. Now that we’ve cleared that fumble, what’s up with all of these different answers people get when doing the same calculations from different reference frames?

Your question was:With respect to your “stationary” observer, each is moving 1/2c. Or are you asking what their apparent difference in velocity is from the view of our hypothetical privlidged observer? In that case, yes, it’s c, but this is a meaningless quantity.
[/quote]

No, that’s fine, that’s all I was wondering about. I know you can see seeming “velocities” of well over the speed of light–for example, the stars appear, in a sense, to move across the sky at many thousands and millions of times the speed of light–without this implying anything is “really” moving that fast. I was just making sure that this concept applies to the situation I was describing as well.

-FrL-