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moebius
reply to post by swanne
In your example the events happen simultaneously in the fixed observer frame of reference. For the observer in the moving train, they won't happen at the same time
Here is the diagram again.
swanne
Arbitrageur
As I already explained, both time and length are affected (see the Minkowski diagram I posted above)
But... to work, this diagram supposes a preferred frame of reference. Otherwise, how can the guy in the train be sure that HE is the one moving in the first place?
It's no illusion. Either perspective is equally real. We humans have some difficulty wrapping our mind around this concept which I guess is why it took centuries for relativity to be discovered. It's not so intuitive.
Because of the train's velocity, it doesn't take 2 seconds for the light from Event B to reach the observer, but in fact 1.3 second (d/(c+v) = 2/(1+0.5) = 1.3). This creates the illusion that Event B happened before Event A.
One thing I noticed in your math, is not so much a math issue is a lack of clarity on which reference frame you are doing the calculations in. You pick one frame or the other, and the speed of light is the same in each. It's the clock rate (time) and length (distance) that's different in the different frames, not the speed of light. If you come up with a different speed of light in the two different frames, you did the math wrong.
swanne
That's because for non-simultaneity to occur, one light ray (the one from Event A) must reach the observer at a slower relative speed than the other ("the other" being the light from Event B). Otherwise both events would be observed as simultaneous again.
When we take the ratios dS/dT = ds/dt = c, we see the only way c can be the same for the outside observer (dS/dT) and the passenger on the train (ds/dt) is for both space and time to adjust to make it so. And that's what happens.
See this link and scroll down to the section on "The Relativity of Time, Mass, and Length", and read the section about length, which hopefully explains how the length is perceived
swanne
Okay, so Light does travel at constant speed, no matter what frame, after all (for this to work, though, it seems length behind the moving observer gets dilated, right? ).
You're welcome and I'm glad you're on the path to figuring this out. If it was easy, it wouldn't have taken centuries after Newton for someone to figure out. It's not easy or intuitive.
I apologize since it took me all that time to get it. You did provide the diagram in Page 1... I didn't realize what you meant... My mistake. Sorry.
My sincere Thanks for your patience.
That's actually a good question. The lightning bolts do not stay locked to the stationary frame. The lightning bolts are specific events in spacetime, and as such their positions do not change in a spacetime diagram when the frame of reference is changed. That's why we use spacetime diagrams.
swanne
One last question, unrelated to the OP topic.
Since no frame can be taken as a preferred reference frame, then why does the Events (lightning bolts) position in both frames stays locked to the Stationary frame, as such:
Events A, B, and C occur in different order depending on the motion of the observer. The white line represents a plane of simultaneity being moved from the past to the future.
It's not the distance from Earth that's changing, it's the position. If a clock is on the ground in London, and the satellite clock travels overhead from New York to Moscow, there is relative motion.
GargIndia
When you place a clock on a satellite, the distance between that satellite and earth is essentially fixed. There is no movement of satellite relative to earth (it is not coming closer or going farther from earth). So relativistic effect should not apply. Why clock on a satellite is showing different time from the one on earth? Please answer this question.
Daily time dilation (gain or loss if negative) in microseconds as a function of (circular) orbit radius r = rs/re, where rs is satellite orbit radius and re is the equatorial Earth radius. At r ≈ 1.497 [Note 1] there is no time dilation. Here the effects of motion and reduced gravity cancel. IIS astronauts fly below, whereas GPS and Geostationary satellites fly above.
GargIndia
reply to post by netbound
My dear friend, you forget that we take measurements of all parameters. The field of measurement is called 'metrology'.
We measure time using some method. We have built clocks for that purpose. There are several types of clocks, each type using a specific method.
When you say 'time dilation' occurs on a moving object wrt to a stationary object, the difference is being shown/computed as difference between two clocks present in these two objects.
We have assumed an atomic clock to be precise whether in a 'Stationary' or 'moving' object? This is an assumption that needs revisiting.
When you place a clock on a satellite, the distance between that satellite and earth is essentially fixed. There is no movement of satellite relative to earth (it is not coming closer or going farther from earth). So relativistic effect should not apply. Why clock on a satellite is showing different time from the one on earth? Please answer this question.
Why clock on a satellite is showing different time from the one on earth?
I said clock speed is related to gravity.
KrzYma
light speed is not constant ! it is related to gravity.
... I see Arbitrageur already explained that
Arbitrageur
I said clock speed is related to gravity.
KrzYma
light speed is not constant ! it is related to gravity.
... I see Arbitrageur already explained that
Light speed is constant, and does not vary, even in varying gravitational fields (according to theory and observations).
The frequency (or wavelength) of light does vary due to gravitational fields however, which we call "red-shift" or "blue-shift".
Energy bends light too but unless you can find an example to the contrary (and I doubt you can), it's negligible. I'll show you how to do the math for the sun.
GargIndia
reply to post by pauljs75
"Light bending around the sun"
Are you sure it is due to 'gravity'.
The stars and interstellar clouds represent regions of high energy in space. It is possible that light is affected by these regions.
I didn't limit the calculation above to just light. By using the mass conversion rate, it should account for all the energy produced by the sun (excepting blips like solar flares/CMEs, or other temporary variations from the norm) even forms of energy beyond the visible spectrum of light.
Yes, in principle but not in any practical sense. The fact that mass can bend light is a well calculated and well measured phenomenon. The first example occurred soon after Einstein developed the general theory of relativity. The measured deviation of distant light from stars was measured during a total eclipse of the sun. The bending angle is proportional to mass of the object causing it. The mass of the sun is enormous but the measured angular deviation was only about 1.5 arc seconds, 0.0004 degree.
The effective mass of any light source that I can think of would not even come close to a microgram of equivalent mass energy. No way can you measure the corresponding deflection it would cause a light beam. Still it's an interesting question, just don't hold your breath.
We certainly have a lot to learn about space (like the nature of dark energy or vacuum energy), but the bigger mystery regarding light getting bent is, what is the source of "dark matter" which seems to bend light more than the visible objects can account for. Until we solve that mystery, our understanding about what exactly is bending the light is incomplete.
'Space' is an item which is least explored and understood by humans. So no need to hurry up until all variables are in place.
The lightning bolts are specific events in spacetime, and as such their positions do not change in a spacetime diagram when the frame of reference is changed.
Now when you draw the Minkowski spacetime diagrams which consider two reference frames, normally you pick one that has right angles of time versus distance, then the other reference frame will NOT have right angles between them. The selection of which one has right angles is arbitrary.
In fact, here is a good illustration showing how three reference frames can be shown on the same spacetime diagram. You see events A B and C simultaneously from one reference frame (say an observer on the ground), then the other two frames could represent trains traveling in opposite directions, one at +0.3c and the other at -0.5c:
Relativity of simultaneity
Events A, B, and C occur in different order depending on the motion of the observer. The white line represents a plane of simultaneity being moved from the past to the future.
That's from the link in the first post in this thread I made, but I don't know if you saw it so I figured I'd repost it since if you can figure out what's going on in that animation it should answer your question.
Similarly, you could redraw the gif animation above for the other two frames and all three will look different, but they will all be mathematically equivalent. The reference frame that's shown with right angles is arbitrary.
It's not easy or intuitive.