Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: miniguy on 26/09/2018 19:25:58
-
Hi,
I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?
-
Hi,
I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?
This is the Doppler effect, the same thing that causes red and blue shift. The faster a thing moves away from an observer, the slower it appears to go, even without time dilation. The effect is much stronger than dilation.
So if a clock is coming at you fast (say a big digital clock on the nose of a ship heading this way), the clock will be slower due to dilation, but still appear to run faster due to blue shift. The clock cannot move at light speed, but it can go as close as you want to it. If you want it to appear to run 10x fast, there is a speed at which that occurs.
So take a clock coming from 8.66 light years away (Earth frame), moving at .866 c, a 10 year trip in our frame. Its clock says 0 when it leaves. At that speed, it will be dilated to half speed and only log 5 years, so it reads 5 years when it gets here. But it leaves 10 years before it gets here, and we don't see the light from its departure for 8.66 years, leaving us watching the clock appear to tick 5 years in only 1.34 remaining years, so it appears to run about 3.73 the rate of one of our own clocks.
Go even faster, and the clock coming at us appears to run even faster, but it is meaningless to posit a clock moving at actual light speed.
-
Watching the clock departing it appears to run slower.
Watching the clock approaching it appears to run faster.
For the duration of the trip, the moving clock accumulated 10 yr in 20 yr earth time.
Observation (including calculations) concludes the moving clock ran slower than the earth clock.
-
So take a clock coming from 8.66 light years away (Earth frame), moving at .866 c, a 10 year trip in our frame. Its clock says 0 when it leaves. At that speed, it will be dilated to half speed and only log 5 years, so it reads 5 years when it gets here. But it leaves 10 years before it gets here, and we don't see the light from its departure for 8.66 years, leaving us watching the clock appear to tick 5 years in only 1.34 remaining years, so it appears to run about 3.73 the rate of one of our own clocks.
Good explanation, Halc. There are a few points/questions that come to mind.
1. The clock, presumably on a craft, travels from A to B. We identify B as Earth.
2. The distance from A to Earth is (presumably) already known, so we can calculate that a craft travelling at 0.866c will take 10 years in Earth’s RF, to make the journey.
3. Light takes 8.66 years to make the same journey.
4. We, on Earth, have no way of knowing when the craft left A. Nor do we have any way of synchronising our clock with that on the craft at the time of its departure.
5. On arrival, would the clock on the craft be 5hrs ahead of clocks on Earth?
6. If the answer to 5 is “yes”, why would that be the case if the clocks were not synchronised when the craft left A.
7. If the craft’s clock differs from Earth’s clock, why would it need “to tick 5 years in only 1.34 remaining years”? Would the difference not remain?
-
Good explanation, Halc. There are a few points/questions that come to mind.
1. The clock, presumably on a craft, travels from A to B. We identify B as Earth.
2. The distance from A to Earth is (presumably) already known, so we can calculate that a craft travelling at 0.866c will take 10 years in Earth’s RF, to make the journey.
3. Light takes 8.66 years to make the same journey.
4. We, on Earth, have no way of knowing when the craft left A. Nor do we have any way of synchronising our clock with that on the craft at the time of its departure.
5. On arrival, would the clock on the craft be 5hrs ahead of clocks on Earth?
6. If the answer to 5 is “yes”, why would that be the case if the clocks were not synchronised when the craft left A.
7. If the craft’s clock differs from Earth’s clock, why would it need “to tick 5 years in only 1.34 remaining years”? Would the difference not remain?
2: Yes, I said that the departure (A) was 8.66 LY away.
4: Well, we're watching. At year 8.66, we see the ship depart abruptly already at speed. Yes, until then, we either don't know, or maybe it was a scheduled thing. We very much can synchronize our clocks, but we need to assume a frame to do it. We assume that A is stationary relative to Earth, and thus that synchronized clocks have meaning. The ship frame is different of course.
5: No... On arrival, the ship clock is 5 years slow. It reads 5, and the Earth clock reads 10, assuming both Earth and 'A' read 0 at the same time in their mutual frame.
7: The OP asks how that clock would appear as it approaches rapidly. We see a zero on the ship clock only 1.34 years before it arrives at time 10. In that 1.34 years, we see it count from 0 to 5 years, and thus 'appears' to run fast when in fact it is dilated by a factor of 2 and only counts 5 years in a journey that actually takes 10 years in Earth's frame.
-
….. At year 8.66, we see the ship depart abruptly already at speed.
Why would it be “already at speed”? If light from “A” could bring us information about one point in time, would it not provide information continuously, such that we would see the process of lift-off?
We very much can synchronize our clocks, but we need to assume a frame to do it. We assume that A is stationary relative to Earth, and thus that synchronized clocks have meaning. The ship frame is different of course.
Point taken. I was confusing clock on craft with clock at “A”.
No... On arrival, the ship clock is 5 years slow
Oops! My bad. I blame my age. :)
Just checking that I have this right. If Earth’s clock and “A’s” clocks are synchronised, then the ship’s clock will be synchronised with both, until it moves, relative to both, but not after that point.
-
Because I'm using SR mathematics, and it does not cover gradual acceleration. GR formulas would need to be used. The OP asked for how a moving clock would appear, but not about a clock that is changing velocity (accelerating). So we either posit an abrupt speed assumed by the clock-displaying ship, or we assumed it was always going that fast and it just happened to read zero on its clock when it passed this point 'A' that is 8.66 light years from Earth.
SR can handle acceleration, the only reason to bring in GR is if you are including gravity in the scenario.
For example:
If T is time as measured by the accelerating ship, t is time as measured in the inertial frame and a is the proper acceleration measured by the ship.
Then, assuming that the ship starts its acceleration at rest with respect to the inertial frame:
t = (c/a) sinh(aT/c)*
T=(c/a) asin(at/c)
v= c tanh(aT/c) = at/sqrt(1+(at/c)^2)
gamma = cosh(aT/c) = sqrt(1+(at/c)^2)
These are all derived from SR without invoking GR.
Granted, it is tricky to work out what is happening to clocks in the inertial frame according to the accelerated frame. (in this case, there is an additional factor that cause clocks you are accelerating towards to run fast by a factor that depends on the magnitude of your acceleration and their distance, and clocks you are accelerating away from to run slow.
* Here t is the time that our inertial frame clock would measure while T would be the time on the ship clock according to the inertial frame. This can also be used to determine what clocks in the inertial frame would read at time T according to the Ship.
So for example, if the ship is accelerating from point A to point B, with clocks at these points which are synchronized in the inertial frame, then, if the ship leaves point A at t=0 and T=0, and the ship reaches B when the ship clock reads T, then t will be the time at B according to the ship upon its arrival at B. The time at A (according to the ship) will depend on the ship's velocity with respect to A and B and the proper distance between A and B( relativity of simultaneity). tB will be greater than T, but tA will be less than T.
-
SR can handle acceleration, the only reason to bring in GR is if you are including gravity in the scenario.
Right you are. I was mistaken to suggest SR doesn't handle acceleration. Gravity is not included, nor is non-local scales. Anything within the galaxy is reasonably local enough for SR.
I was keeping it simple to answer the simple question asked in the OP: Does an approaching clock appear to run faster, and yes, it does.
So for example, if the ship is accelerating from point A to point B, with clocks at these points which are synchronized in the inertial frame, then, if the ship leaves point A at t=0 and T=0, and the ship reaches B when the ship clock reads T, then t will be the time at B according to the ship upon its arrival at B. The time at A (according to the ship) will depend on the ship's velocity with respect to A and B and the proper distance between A and B( relativity of simultaneity). tB will be greater than T, but tA will be less than T.
I take it that tA is what A clock currently reads in ship frame. So in my simplified example where the ship is already moving when it departs A, T at B is 5, tB is 10, and tA is 2.5. Yes, I agree with all this given that the acceleration vector is forwardish (scientific term!) the whole way. If the ship turns around and accelerates the other way for part of the trip, or maybe makes a detour to Arcturus for nachos, it arrives at B with tA possibly more than T, and tA and tB both synced if the ship velocity becomes zero in the inertial frame of A and B. So tA is less than T only if acceleration is predominantly away from A.
-
SR can handle acceleration, the only reason to bring in GR is if you are including gravity in the scenario.
Right you are. I was mistaken to suggest SR doesn't handle acceleration. Gravity is not included, nor is non-local scales. Anything within the galaxy is reasonably local enough for SR.
I was keeping it simple to answer the simple question asked in the OP: Does an approaching clock appear to run faster, and yes, it does.
So for example, if the ship is accelerating from point A to point B, with clocks at these points which are synchronized in the inertial frame, then, if the ship leaves point A at t=0 and T=0, and the ship reaches B when the ship clock reads T, then t will be the time at B according to the ship upon its arrival at B. The time at A (according to the ship) will depend on the ship's velocity with respect to A and B and the proper distance between A and B( relativity of simultaneity). tB will be greater than T, but tA will be less than T.
I take it that tA is what A clock currently reads in ship frame. So in my simplified example where the ship is already moving when it departs A, T at B is 5, tB is 10, and tA is 2.5. Yes, I agree with all this given that the acceleration vector is forwardish (scientific term!) the whole way. If the ship turns around and accelerates the other way for part of the trip, or maybe makes a detour to Arcturus for nachos, it arrives at B with tA possibly more than T, and tA and tB both synced if the ship velocity becomes zero in the inertial frame of A and B. So tA is less than T only if acceleration is predominantly away from A.
Right. So for example, if the ship were to suddenly accelerate to come to rest with respect to A and B upon reaching B, then tA will advance to equal tB upon completion of the acceleration.SR can handle acceleration, the only reason to bring in GR is if you are including gravity in the scenario.
Right you are. I was mistaken to suggest SR doesn't handle acceleration. Gravity is not included, nor is non-local scales. Anything within the galaxy is reasonably local enough for SR.
I was keeping it simple to answer the simple question asked in the OP: Does an approaching clock appear to run faster, and yes, it does.
So for example, if the ship is accelerating from point A to point B, with clocks at these points which are synchronized in the inertial frame, then, if the ship leaves point A at t=0 and T=0, and the ship reaches B when the ship clock reads T, then t will be the time at B according to the ship upon its arrival at B. The time at A (according to the ship) will depend on the ship's velocity with respect to A and B and the proper distance between A and B( relativity of simultaneity). tB will be greater than T, but tA will be less than T.
I take it that tA is what A clock currently reads in ship frame. So in my simplified example where the ship is already moving when it departs A, T at B is 5, tB is 10, and tA is 2.5. Yes, I agree with all this given that the acceleration vector is forwardish (scientific term!) the whole way. If the ship turns around and accelerates the other way for part of the trip, or maybe makes a detour to Arcturus for nachos, it arrives at B with tA possibly more than T, and tA and tB both synced if the ship velocity becomes zero in the inertial frame of A and B. So tA is less than T only if acceleration is predominantly away from A.
So for example, if the ship were to accelerate at a to the midpoint of A and B, flip around, maintain the same proper acceleration in opposite direction so that it comes to rest next to B.
Then:
At the midpoint, tA will be less than T and Tb will be greater than T as measured from the ship.
Upon arrival at B, tA=tB and will be greater than T.
-
The clock on the craft might read zero at event A, but it is in sync with no other clock since it is not stationary relative to any other clock. So it can be synced to an event, but not to any clock.
It is instructive to describe the entire situation from the POV of the ship. What does Earth's clock appear to read if the ship observer is watching it during the entire time between events A and B? Clue: It obviously reads 10 at event B when the ship is in that clock's presence.
I’m labouring points a bit, but I like to feel sure I’m on the right track and can cope with the next “what if…” that comes into my head.
Do I have this right?
If we consider Earth and A to be stationary relative to each other; their clocks can be synced at t=0.
If, at t=0, the ship is stationary relative to A and its clock is synced with A’s clock; the ship’s clock will be synced with Earth’s.
However, if we start the scenario with the ship passing A at 0.866c (or any speed); there will be an instant when the clocks on A and the ship can be synchronised (this we call t=0), but at that point the ship’s clock will not be synced with Earth’s.
Who wanted nachos, anyway? :)
-
If we consider Earth and A to be stationary relative to each other; their clocks can be synced at t=0.
Synced in the frame in which they are stationary, yes. Not synced in other frames.
If, at t=0, the ship is stationary relative to A and its clock is synced with A’s clock; the ship’s clock will be synced with Earth’s.
However, if we start the scenario with the ship passing A at 0.866c (or any speed); there will be an instant when the clocks on A and the ship can be synchronised (this we call t=0), but at that point the ship’s clock will not be synced with Earth’s.
No, in the frame of the moving ship, the clock at A is not in sync with the clock at B. So you can say that there is that instant when the ship clock is synced to the departure event at B (synced to an event, not to a clock). Both clocks are present at that event and they both read zero (are set to zero actually) at that event.
As for the ship parked before departure, it is stationary in B’s frame, not the eventual frame of the moving ship.
-
Synced in the frame in which they are stationary, yes. Not synced in other frames.
Great – we can lay that one to rest!
No, in the frame of the moving ship, the clock at A is not in sync with the clock at B.
I have no problem with that, but I was trying to establish whether:
If, at t=0, the ship is stationary relative to A and its clock is synced with A’s clock; the ship’s clock will be synced with Earth’s.
At this point the ship would not be moving relative to A or B, so would its clock be synced with those of A and B?
As for the ship parked before departure, it is stationary in B’s frame, not the eventual frame of the moving ship.
Possibly the answer is here and I’m missing it. If “it” is the ship, it is stationary relative to B, that’s fine, but what are you saying is not “stationary” in “the eventual frame of the moving ship”?
-
If, at t=0, the ship is stationary relative to A and its clock is synced with A’s clock; the ship’s clock will be synced with Earth’s.
At this point the ship would not be moving relative to A or B, so would its clock be synced with those of A and B?
In the frame where A and B is stationary, and so is the ship still, all three clocks are synced, yes.
As for the ship parked before departure, it is stationary in B’s frame, not the eventual frame of the moving ship.
Possibly the answer is here and I’m missing it. If “it” is the ship, it is stationary relative to B, that’s fine, but what are you saying is not “stationary” in “the eventual frame of the moving ship”?
Well, the ship isn’t going to stay at B is it? It is going to accelerate and head for A (home presumably), and that makes it stationary in a new different frame in which clocks A and B are not synced.
-
Well, the ship isn’t going to stay at B is it? It is going to accelerate and head for A (home presumably), and that makes it stationary in a new different frame in which clocks A and B are not synced.
You’re ahead of me there, I had not considered the return trip.
As mentioned in another thread, some years ago I put quite a lot of time and effort into sorting out in my mind a similar scenario involving three “moving” ships. A & B were maintaining a constant separation and C was moving relative to them. Somewhere in the recesses of my brain I must have had the idea that because A & B, in this case, were planets the situation would be different.
Thanks for helping me to exorcise that one.
-
Well, the ship isn’t going to stay at B is it? It is going to accelerate and head for A (home presumably), and that makes it stationary in a new different frame in which clocks A and B are not synced.
You’re ahead of me there, I had not considered the return trip.
My apologies for getting A and B mixed up. Your scenario was starting at A and going to B, with B being Earth, as per your first post in this topic. Kindly switch A and B in my quotes where I got that backwards. Duh… Of course A then B. Events are best labeled in alphabetical order as you’ve done.
As mentioned in another thread, some years ago I put quite a lot of time and effort into sorting out in my mind a similar scenario involving three “moving” ships. A & B were maintaining a constant separation and C was moving relative to them. Somewhere in the recesses of my brain I must have had the idea that because A & B, in this case, were planets the situation would be different.
Doesn’t matter if any of them were ships or planets or just clocks. Pretty hard to get a ship up to .866 c but not so hard to blast clocks out of a gun.
They actually have crude clocks that move at well over 99% light speed fired at the observer from say 100 km away, set to self-destruct after a mean of say 2.2 microseconds (time to move about 2/3 km) but it takes light about 3000 microseconds to go that distance so the vast majority of the ‘clocks’ should have self-destructed before that distance is covered, but most of them make it due to time dilation. This is the muon experiment. Muons and other decaying particles make pretty good clocks. Same principle used in carbon dating.
-
Hi,
I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?
Nothing can travel "at light speed", so you must have heard wrong or the source is wrong.
-
https://physics.stackexchange.com/questions/194921/if-traveling-at-the-speed-of-lights-stops-time-why-does-it-take-light-8-minutes
The thing is, in relativity you cannot have a reference frame "chasing" a photon. You'll get singularities if you try to view the world from a photon's perspective. A photon cannot move like you and you cannot move like a photon.
As a photon, travelling along a light-like world line, experiences no proper time it's proper velocity is simply undefined. However, we can perfectly state that the speed of light is invariant for every reference frame where v<cv<c.
Special relativity breaks down when going extreme (e.g. for velocities equal to or higher than cc), but the absence of tachyons is also an experimental fact. There are probably no such things either.
All models breaks down at some point, but as long as it produces good pragmatic experimentally results, the model is useful within the set boundaries. The proper velocity of light is simply not well-defined; it's meaningless to talk about a proper velocity for a photon nor a valid reference frame in which a photon is at rest. Such concepts have no defined meaning and lies outside what the model (in this case, special relativity) can deal with.
What a scientist does when a model gives you nonsense is to go back and ask nature for hints for a better model. We do already know that the theories of relativity is incomplete, and we'll never achieve a complete theory. We can, however, do our best to continue our investigation and come up with models as complete as we can make them.
Seems like a reasonable approach.
-
Hi,
I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?
Nothing can travel "at light speed", so you must have heard wrong or the source is wrong.
I wouldn't go so far as to say that they heard wrong or that the source is wrong.
It is quite possible that Einstein considered such a thought experiment. And he might have even considered the "frozen clock" observation. The point is that he finally concluded that the you couldn't formulate a consistent picture of what such an observer would see, and that such a scenario could not arise in reality.
-
Hi,
I watched a video on Youtube about an Einstein thought experiment, in which he realised that if he was travelling away from a clock at light speed the clock would appear to be frozen, but what happens if you travel at the speed of light towards the clock - does the clock appear to speed up?
Nothing can travel "at light speed", so you must have heard wrong or the source is wrong.
I wouldn't go so far as to say that they heard wrong or that the source is wrong.
It is quite possible that Einstein considered such a thought experiment. And he might have even considered the "frozen clock" observation. The point is that he finally concluded that the you couldn't formulate a consistent picture of what such an observer would see, and that such a scenario could not arise in reality.
And we're talking about SR in its final form, not during his musings about it. I.e. we're speaking about mainstream - current - physics. That's why I wrote what I did.