According to Einstein's Theory of Relativity, massive bodies cannot travel at or faster than the speed of light (c), but they can approach it arbitrarily closely. Massless particles, such as photons, gluons, and theorized gravitons, always travel at c. The discussion emphasizes the relativity of time and distance, indicating that while a traveler may perceive a shorter journey time, the coordinate time for an observer at rest remains longer. The conversation highlights the importance of understanding relativistic effects when discussing high-speed travel.
PREREQUISITESStudents of physics, educators, and anyone interested in the implications of relativity on high-speed travel and the nature of light and mass.
Seriously... get a grip! You are talking about the time it takes light to go from Earth to Sirius B as measured on Earth clocks! Distance and time are frame-dependent quantities.MeJennifer said:
Didn't you say 8.6 before? That's also what it says in the chart at the bottom of this article.MeJennifer said:
If Sirius B is 8.6 ly away as you said, Earth's clock will show that 17.2 years pass between when the light is sent out and when it returns. But do you really think that other frames cannot explain this perfectly well in terms of their own coordinates? Take the frame of someone moving at 0.6c relative to Earth/Sirius B, so the Lorentz contraction and time dilation factor 1/gamma is 0.8. In their frame, the distance between Earth and Sirius B is 8.6 * 0.8 = 6.88 light-years. If they see Sirius B and Earth moving in the +x direction, with Earth at x=0 when the light is emitted and Sirius B at x=6.88 at the same moment, then you have to remember it will take longer than 6.88 years in this frame for the light to reach Sirius B, since Sirius is rushing away from it at 0.6c. To find the time, just solve for t in 1c*t = 0.6c*t + 6.88 light years (because the photon's position as a function of time is x(t) = 1c*t, while Sirius B's position as a function of time is x(t) = 0.6c*t + 6.88 ly, so setting them equal shows when they meet), which gives t = 6.88/0.4 = 17.2 years. Then when the light is reflected back, it takes less than 6.88 years for the light to return to Earth, since the Earth is rushing towards the point it was reflected; the time can be found by solving for t in 6.88 ly - 1c*t = 0.6c*t which gives t = 6.88/1.6 = 4.3 years. So, in this frame the total time for the light to go from Earth to Sirius B and back is 17.2 + 4.3 = 21.5 years. But don't forget that in this frame, Earth's clocks are slowed down by a factor of 0.8, so during this time they have only advanced forward by 21.5 * 0.8 = 17.2 years, exactly the same prediction as was made in Earth's frame.MeJennifer said:
That may be so by his clock but he is not directly measuring the rountrip time at all as that would be impossible. Fact remains that it takes 17.2 years for light to make a roundtrip from Earth to Sirius B.JesseM said:
How else do you propose to make a direct measurement on the rountrip time of light from the Earth to Sirus B?Doc Al said:
OK, so imagine a clock that starts at Earth at the moment the light is emitted, accelerates to some high fraction of lightspeed, then later turns around and makes it back to Earth in time for the light's return. Here we have used a single clock to measure both departure and return too, so isn't this a "roundtrip" time? If you restrict roundtrip time to inertial clocks, what physical justification do you have for saying roundtrip time measured by inertial clocks is somehow more "real" than roundtrip time measured by non-inertial clocks?MeJennifer said:
You are missing the point, the traveling clock will not measure the roundtrip time of light at all it will simply desynchronize from the Earth's clock due to its speed differential when it traveled and hence all it will "measure" is the time difference between the Earth's clock when it will come back to Earth.JesseM said:
Do you call it "traveling" in contrast to the Earth clock? But of course there are countless frames where the Earth clock is traveling too. There is no physical sense in which the time between two events as measured by an inertial clock that has the two events on its worldline is any more "real" than the time between the same two events as measured by a non-inertial clock that has the events on its worldline; they are just two different proper times for two different clocks.MeJennifer said:
The only physical notion of time in relativity is proper time. Again, one clock's proper time isn't any more "real" than any other clock's proper time. So although you can talk about the proper time for light to go from Earth to Sirius B and back as measured by a clock on Earth, there's no reason to think of this as "the" time for light to go from Earth to Sirius B and back.MeJennifer said:
Funny thing about probabilities is that it is exactly as likely to happen tomorrow as on the last day of the universe.peter0302 said:
peter0302 said:
A strawman argument as I am not disagreeing with that at all.JesseM said:
There is only one proper time between any sequence of events or in this case 3 events, the emission, relfection and absorption of light. All observers must agree on the elapsed proper time between those events. That such an amount of elapsed proper time does not agree with their clocks is either due to relative motion or spacetime curvature.JesseM said:
Really? Then why did you place so much emphasis on this statement:MeJennifer said:
If you agree that light only takes 2*8.6 years round trip according to one arbitrary clock, and that the time would be different according to other equally valid clocks, then why do you think Sirius B "is" 8.6 light years from Earth? Do you agree that there is absolutely nothing physically special about the measurements that give the round-trip time as 2*8.6 years as opposed to some other number?
There is only one proper time on a given worldline that contains those events, but there are multiple possible worldlines that contain the events, and they measure different proper times between the events. Do you disagree?MeJennifer said:
Do you disagree that in relativity the term "proper time" is only used in the context of particular worldlines, that there is no unique "proper time" between distinct events which can have multiple worldlines that pass through both? If you do disagree, then you are using the term "proper time" incorrectly.MeJennifer said:
"Relative motion" relative to what exactly? You certainly can't talk about motion relative to the events themselves, since events are instantaneous...MeJennifer said:
Do you realize that there is only one possible worldline (or multiple in certain curved spacetimes) between events that are lightlike separated?JesseM said:
Sure, but the proper time along a null geodesic is always 0 (or maybe physicists don't even talk about 'proper time' for null geodesics, but in the limit as timelike geodesics get closer and closer to null geodesics the proper time should approach 0).MeJennifer said:
JesseM said:
Yes, but as I said in response to the earlier comment you edited, if it makes sense to talk about proper time on this worldline at all, the proper time would be zero. So this still doesn't help make sense of your comment that the time between light leaving Earth and arriving at Sirius B "is" 8.6 years.MeJennifer said:
Yes, in the Earth frame the distance between the traveler and the photon is only increasing at a rate of 0.1 light-seconds per second.spiffomatic64 said:
No, the Earth will measure the traveler's clocks to be slowed down rather than speeded up (and will also measure the traveler's rulers to be shrunk in the direction of motion). In relativity, any observer moving inertially (constant velocity) will measure clocks that are moving relative to themselves to be running slow.spiffomatic said:
That is the proper time of the photon but alternatively we can use several affine parameters for a null geodesic.JesseM said:
But the affine parameters aren't really "time". Anyway, are you claiming that when you said the time "is" 8.6 years and the distance "is" 8.6 light-years, you were thinking in terms of affine parameters on the photon's worldline? If not, how is this relevant to what we were talking about before?MeJennifer said:
And couldn't you come up with different affine parameters for the same photon worldline based on travel time in different frames? There wouldn't be a unique choice of parameter forced on you by physics like there is with proper time on timelike worldlines.MeJennifer said:
What you say does not make any sense to me. Affine parameters operate on spacetime curves while frames are 3D hypersurfaces of spacetime.JesseM said:
You can parametrize a spacetime curve in terms of the time-coordinate assigned to each event on that curve by a particular frame, no? Of course I don't know if this parameter would qualify as an "affine" parameter since I'm not too familiar with GR. But what did you mean when you said "one could for instance use 'travel time' as an affine parameter"? Travel time according to what coordinate system or clock?MeJennifer said:
The answer is no, think of a common point and the light traveling in the form of an expanding sphere then it is guaranteed that the massive object is always inside this sphere. If for the sake of argument that massive object were to be found outside the sphere then the spacetime causal structure would be violated under the constraints of GR.matheinste said:
The question is a little ambiguous. In GR it is possible that if a particular photon departs from a massive object, that particular photon will take longer than the massive object to reach some other destination (think of photon orbits around a black hole, and imagine the massive object and the photon departing in opposite directions, so that the object has a short distance to reach some nearby buoy that we label as the destination, while the photon has to go all the way around the black hole before it hits the buoy). Still, in this situation it should always be possible to imagine a different photon which departs from the same point in spacetime but in a different direction, and which reaches the destination before the massive object.matheinste said:
All you really need to know is that at a certain distance from the event horizon, a photon released at the right angle will orbit in a circle around the black hole. And if you have a massive object and a photon going in opposite directions, and there's a buoy nearby in the direction that the massive object is going, the massive object could reach it before the photon, since the photon is making a longer trip all around the black hole.matheinste said:
As long as the photon is moving unimpeded (no mirrors to reflect it, for example), the photon will always reach a given destination before the massive object.matheinste said: