Thanks for taking that in the right spirit. I should use those little smiley’s more often to indicate when I’m not being too serious.Ice cream also has nothing to do with special relativity.
Relativity - Oh dear, here we go again!
- Thread starter ynot
- Start date
Do particle accelerator actually work? How’s that hugely expensive hadron collider coming along?
<<< NOTEI meant what was the original problem(s) not how it has been used since.
Reality Check
Penultimate Amazing
Read History of special relativity.Do particle accelerator actually work? How’s that hugely expensive hadron collider coming along?
I meant what was the original problem(s) not how it has been used since.
The basic problem that Lorentz, etc. worked on for many years was that Maxwell's equations do not look the same for moving observers. Note that Einstein's 1905 paper is actually about the electrodynamics of moving bodies.
sol invictus
Philosopher
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It's not just a question of appearances - one twin comes back younger.
Perhaps a better analogy is to think of odometers attached to two cars (this analogy relates the odometer to a clock: one measures distance traveled through space, and the other "distance" traveled through time). The two cars can leave some location, travel on two different paths, and return to the same point. If the two paths are of different lengths, the cars will come back with two different odometer readings - one car is "older" than the other.
What Einstein taught us is that the same type of phenomenon occurs for travel through time as well. Not only will the odometers of the two cars read differently, the clocks they carry will as well. That experiment has been done - there is no question as to its validity (you could try to argue that relativity is not the explanation, but you cannot deny that moving clocks run slow). In fact these effects are observed routinely in satellites, for example in the GPS system, as well as in airplanes. Atomic clocks are far more accurate than is necessary to detect this effect (which is small, but not that small).
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Ok, that's clearer.
She still travelled forward through time to the moment where the other twin returned.
Hmm, this should probably be adressed by one of the more knowledgeable posters, but I will try to answer it.
If you simply measure the distance between the starting point and end point of their movements, A and B obviously have travelled the same distance. But if you measure the actual lengths of the paths they travel (so that 1m forward + 1m back = 2m), then A:s path will be longer. This should be true as long as any inertial coordinate system is used, even in Newtonian mechanincs.
The same thing happens to the twins in special relativity. Measuring the lengths of the paths they travel through space-time will show you that the twin in the space-ship takes a path which is shorter in the time-dimension, meaning that she will be younger. Note that all inertial coordinate systems will agree on this, even if they will disagree on the distance between the starting-point (spaceship leaves) and end-point (spaceship returns) of the twin's journey.
Fredrik
Graduate Poster
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About that conveyor belt...
Let's choose the inertial coordinate system in which the conveyor belt is at rest and B starts moving in the positive x direction with speed u at the event with coordinates (t,x)=(0,0). The speed of the belt relative to the ground is irrelevant (assuming it's constant).
What's A's path through spacetime during the next T seconds on his own clock? It's the path that we would describe as the straight line from (0,0) to (T,0) if we use this particular coordinate system. What's B's path through spacetime between the same two events? It's the one that we would describe as consisting of two straight line segments, the first from (0,0) to (T/2,uT/2) and the second from (T/2,uT/2) to (T,0), if we use the same coordinate system again.
The proper time of any path (through spacetime) between these two events is
[latex]\int\sqrt{dt^2-dx^2}=\int_0^T dt\sqrt{1-\frac{dx^2}{dt^2}}=\int_0^T dt\sqrt{1-v^2}[/latex]
where t and x are the temporal and spatial coordinates of the coordinate system discussed above. The proper time integral is completely independent of the coordinates we use (the proof of that involves math), but the calculation is the easiest when we use that particular coordinate system. (That's the only reason we're using it).
For A, we have v=0 from t=0 to t=T. For B, we have v=u from t=0 to t=T/2 and v=-u from t=T/2 to t=T. So the proper times τA and τB of the paths of A and B are
[latex]\tau_A=T\sqrt{1-0^2}=T[/latex]
[latex]\tau_B=\frac T 2\sqrt{1-u^2}+\frac T 2\sqrt{1-(-u)^2}=T\sqrt{1-u^2}[/latex]
These are the times displayed by their own clocks, and it's also the times by which they have aged between the two events.
Let's choose the inertial coordinate system in which the conveyor belt is at rest and B starts moving in the positive x direction with speed u at the event with coordinates (t,x)=(0,0). The speed of the belt relative to the ground is irrelevant (assuming it's constant).
What's A's path through spacetime during the next T seconds on his own clock? It's the path that we would describe as the straight line from (0,0) to (T,0) if we use this particular coordinate system. What's B's path through spacetime between the same two events? It's the one that we would describe as consisting of two straight line segments, the first from (0,0) to (T/2,uT/2) and the second from (T/2,uT/2) to (T,0), if we use the same coordinate system again.
The proper time of any path (through spacetime) between these two events is
[latex]\int\sqrt{dt^2-dx^2}=\int_0^T dt\sqrt{1-\frac{dx^2}{dt^2}}=\int_0^T dt\sqrt{1-v^2}[/latex]
where t and x are the temporal and spatial coordinates of the coordinate system discussed above. The proper time integral is completely independent of the coordinates we use (the proof of that involves math), but the calculation is the easiest when we use that particular coordinate system. (That's the only reason we're using it).
For A, we have v=0 from t=0 to t=T. For B, we have v=u from t=0 to t=T/2 and v=-u from t=T/2 to t=T. So the proper times τA and τB of the paths of A and B are
[latex]\tau_A=T\sqrt{1-0^2}=T[/latex]
[latex]\tau_B=\frac T 2\sqrt{1-u^2}+\frac T 2\sqrt{1-(-u)^2}=T\sqrt{1-u^2}[/latex]
These are the times displayed by their own clocks, and it's also the times by which they have aged between the two events.
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Dancing David
Penultimate Amazing
Yeah but the rate of time/change is constant until they move in different frames of reference. This sidebar does not effect anything.
Dancing David
Penultimate Amazing
Hmmm, then how do you explain the difference in rates of passge of time for cesium clocks in orbit? That is one of the data points that agrees with time dialation. Clocks in sperate frames of reference do show small time dialation effects relative to clocks on the earth and they are in agreement with the theory.
That validates a building block right there. It demonstrates something that looks like time dialation.
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Dancing David
Penultimate Amazing
Dancing David
Penultimate Amazing
Excuse me?
We are not talking about a correlative measure of .65 here.
We are talking about .99999999 which is much higher.
Science uses approximate models, none of them are 'real'. So by your standard all science is hooey.
The issue is that you just want to assert that you are right.
The approximate model of relativity makes very accurate predictions, it is not in the ice cream causes drowning category. It is in the approximate model category, if you do not like it then you know what the solution is? Come up with a more accurate approximate model that makes a better prediction.
What your argument is saying is exactly the same as saying
"Germ and virus theory of infection could be the product of ongoing subconscious confirmation bias."
"The pancreatic theory of diabetes could be the product of ongoing subconscious confirmation bias."
"The gravitational theory of orbits could be the product of ongoing subconscious confirmation bias."
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Dancing David
Penultimate Amazing
I'm starting to get the impression that ynot doesn't actually want to understand this.
This is not the first time validation of the twin "paradox" using caesium clocks has been bought up. Does he have a short term memory deficit perhaps?
This is not the first time validation of the twin "paradox" using caesium clocks has been bought up. Does he have a short term memory deficit perhaps?
Mashuna
Ovis ex Machina
I believe he's got an identical twin who understood it fine.
Dave Rogers
Bandaged ice that stampedes inexpensively through
Sadly, this seems to be the case. The not understanding relativity part isn't a problem - it's not particularly intuitive and lots of people have trouble with it, especially those not so strong on the maths front. However, questions like this:
are rather different. This thread was started two years ago, almost to the day, yet in all that time ynot hasn't bothered to spend 10 seconds with Google or Wikipedia to find the answer to a simple question, the answer to which is part of most high school physics courses.ynot said:
Ignorance isn't a major problem and can be cured. Deliberate ignorance is something else entirely. Especially when coupled with the astounding arrogance to declare the experts wrong.
