Revisiting Einstein's Train Experiment: Unraveling the Mysteries of Relativity

[GhostLoveScore]

Let's put it this way. Those two light bulbs are blinking. While the train is not moving, they blink at the same time. Then we start moving. What would we see? If I understood this correctly, they would continue to blink at the same time.

 

[Orodruin]

Then we start moving. What would we see?

You are still not specifying the situation properly. You do not mention how the train starts moving. Again, you could specify that the entire train starts moving "at the same time" but this is still ambiguous since you are not specifying which frame you are referring to with your "at the same time". You are also not specifying what you mean by "at the same time" when you state that "they would continue to blink at the same time" - you must specify which frame you are referring to when you make such a statement or the statement will be meaningless.

 

[15characters]

I see, now it's clearer. So in the same situation, train moving to the right - if we have two light bulbs each on one end of the train and we have a switch in the middle of the train where we are. That switch turns on both lights at the same time. What would we see? What would an observer on train station see as we are moving past him?
The difference here is that the light bulbs are moving with the train. The lightning strike was stationary for external observer.

Excellent example. I only just understood how it worked last week. For those on the train and those on the Earth the switching on of the light was an event.

Lets imagine the light bulb in the middle of the train is turned on. For those on the train it is easy to guess what will happen. They are stationary and so is the train. Therefore the light from the light bulb in the middle of the train will hit both end of the train simultaneously.

Now, for you watching from the train track, the train is moving. The back end moves towards the light bulb. The font end moves away from the light bulb. Therefore it is easy to guess what you will see. You will see the light hit the back end first.

How is the phase gap between clocks calculated

Please understand how the difference between the clocks at front and back of train is calculated. It is very simple. Length of train X Relative velocity = time gap.

So the train is 10 light seconds long and moving at 0.5C. What is the gap between front and back clocks?
The Answer is 10 X 0.5 = 5 seconds. It really is that simple to calculate!

If the train was 100 light seconds long and moving at 0.9c what would be the time gap between front an back clocks?

100 x 0.9 = 90 seconds.

Hard question

Moon is 1.25 light seconds away. We have clocks on the moon which we have synchronised with earth.

You fly over Earth towards the moon moving at speed 0.8C. The clock on Earth says the time is 0. What does the clock on the moon say. How many seconds into the future is it?

The answer below

1.25 x 0.8 = 1 second!

Congratulations if you worked it out it is 80% of understanding relativity in my opinion. Learn the invariant equation and you are 100% there. Put it in excel play around converting different times and distances for you and people on the train.

The time gap you see between front and back clocks on a train passing you increases in direct proportion to the proper length of the train and is calculated as length x velocity (for you).

As for your second question, well now you have the formula to see what the time gap between any two clocks on a moving train will be :)

Please note that if you are ON the train the train is NOT moving so there will be no time gap between train clocks.

 

[FactChecker]

Let's put it this way. Those two light bulbs are blinking. While the train is not moving, they blink at the same time. Then we start moving. What would we see? If I understood this correctly, they would continue to blink at the same time.

You have to be more careful when you say "at the same time" for two events that are separated in distance. As @Orodruin has been saying, observers on the train do not agree with stationary observers about "simultaneous" events that occur at a distance. No one person can observe both events since they are separated and moving. Suppose you are on the ground and have carefully synchronized clocks with observers along the tracks. When the light hits the targets, your observers at each target location record the time it hit. They are widely separated but they are using your synchronized clocks. They say that the light hit the back target much earlier than the front target. Suppose people on the train have done the same thing and synchronized their clocks within the train. They record their time at each target location that the light hits both back and front targets. Their recorded times are identical. (In reality, famous experiments came to that conclusion no matter which direction the light was pointed versus the Earth's motion). So you and the train people disagree on what "simultaneous" means for events that are at a distance.

In your example of the light bulbs, the train people would say that the bulbs are always blinking simultaneously. The people on the ground would agree at first, when the train is stationary, but would start to disagree more and more as the train sped up. The rear light would be blinking earlier according to the ground synchronized clocks and the front light would be later.

It's hard to explain why. Experiments kept showing that light traveled at a constant speed in all directions regardless of how fast the person measuring it went in any direction. It's just a fact of physics.

 

[15characters]

Can you than explain why they occur at different times? And please don't say "because the train is moving".

I can follow your thinking cause I was struggling with the same questions a few days ago. Your first question was

"3) If we were moving with 0.99c towards some galaxy. When we would look at it we would see it accelerated because we were moving towards its light and its time would seem to flow faster?"

A) Imagine flying past Earth towards this Galaxy at 99%C. Earth time as you pass is year 0.00

B) The Galaxy is 100 light years away and according to Earth people the time on the Galaxy clock shows year 0.00 at this precise moment Earth clock shows 0.00.

C) You in your ship disagree. Front Clocks run slow by Velocity x Distance. Thus, according to you time on the Galaxy at Earth time = 0.00 is Galaxy time = year 99.00.

(100 light years distance x 0.99c velocity = 99 years of timing phase gap

D For people on Earth your journey to the Galaxy at 0.99c would take 101.01 Earth years. And when you arrive Earth and Galaxy clocks will say 101.01

E) 0.99c produces a gamma factor of 7.09. So for you the distance to the galaxy will be compressed to 14.1 light years and the journey will last 14.2 years.

So:

i) Time on Earth NOW as you pass it is year 0.00. Photons are arriving from the distant Galaxy. They are dated "Galaxy/Earth year -100"

ii) Earth people say the photons now arriving in Earth at speed C from the Galaxy were emitted 100 years ago when the Galaxy clock said year -100. This makes sense to Earth because the Galaxy is always 100 llight years away.

iii) As you fly over Earth at 0.99C the same photon batch labeled "Galaxy/Earth year -100" also hit your rocket ship ...also at speed C. They are pursued by the galaxy rushing in behind yet lagging behind them slightly at a mere 0.99c,

How old are the photons hitting you and the Earth from the Galaxy?

We know the photons from the Galaxy were at one point in time, in the Galaxy. When was this?

If the Galaxy is approaching at 0.99C and the photons at 100%C than they have separated from each other at a speed of 0.01C. At a speed of 0.01 C how long would it take for this photon batch to be 14.1 light away from the galaxy as they are NOW?

14.1 years of distance / speed of 0.01C = time of 1,410 rocket ship years ago.

So 1,410 rocket years X GAMMA is equal to 199.00 Earth years.

Earth says photons are 100 Earth years old we they are 199 Earth years old

Earth says photons arriving now are 100 light years old, and the Galaxy clock must therefore say -100 + 100 = year 0.00

Whereas, we look at the same photons arriving from the galaxy and we say they are ancient and were sent 1,410 rocket ship light years ago when the galaxy was 1,410 light years away by our clocks. We convert the rocket years into Earth years. Earth and the Galaxy clocks run 7 x slower than rocket clocks so 1,410 rocket ship years is 199 Earth years.

What is the current time in the Galaxy based on the photons/images we and Earth are both receiving at Earth time 0.00

Earth says the images it sees from the Galaxy are from 100 light years ago (Earth year -100). We in the rocket say images are from 199 Earth years ago.

So if the images from the Galaxy are from Galaxy clock time -100 years we would expect actual time on the Galaxy NOW to be -100 years plus 199 years. So we say the Galaxy time at Earth time 0.00 is year 99.00

Therefore we and Earth disagree. According to us time now on the Galaxy is year 99.00 . For Earth the time on both Galaxy and Earth clocks is year 0.00.

The difference is 99 years (length x velocity)

So there is a difference of 99 years between what we in our rocket and what those on Earth say the date is NOW in the other Galaxy.

We say what we see NOW is 199 years old and that the Galaxy clock NOW actually shows year 99. Whereas Earth says the images are only 100 light years and so the galaxy clock NOW shows year 0.00.

This is a difference between front and back clocks of 99 years!

Do you know another way to calculate this time difference between Earth and the Galaxy for us.

Distance X Velocity = 99 years!We consider the Galaxy Clock to be 99 Earth years ahead of Earths clock, yet we also claim it is running very slow. How does this work?.

The Galaxy currently at distance of 14.1 light years is moving towards us at 0.99C, It will arrive in 14.2 rocket years. In those 14.2 years the slow Earth / Galaxy clocks will advance by only 2.01 years of Earth / Galaxy time.

When we pass the Galaxy we take a photograph of the Galaxy clock which is synchronised with Earth's clock. What will the Galaxy clocks say at that point?

For Earth

Earth says it is year 0.00 now on the Galaxy. Earth says our trip took 101.01 years. So when we arrive time on Galaxy will be 101.01 years.

For us

Well, according to us Galaxy is at year 99.00 when we pass the Earth clock showing 0.00. Our journey will last 14.2 rocket years. After 14.2 years of our time the slow Galaxy clocks will only move by 2.01 years So when we arrive they will say 101.01 years. So we can take the photo.At the end of the trip

At the end of the trip everyone on the Galaxy both inhabitants and us in our speeding rocket flying past will agree that Galaxy time on our arrival was 101.01. We can take photos.

However, we will disagree on what Earth's clock says at the same moment based on our calculations.

Galaxy inhabitants will say Earth time at our arrival is identical to Galaxy Time 101.01

Rocket people agree Galaxy Time is 101.01 on our arrival. However, we will claim the The Earth Clock simultaneously shows only 2.01. This being 0.00 at our departure plus 2.01 slow Earth years recorded for our trip.

So the time gap of 100 X 0.99C = 99 years remains on arriving in the Galaxy.

Finally

Would we see time accelerate as we approached the galaxy - or rather as it approached us! .

Well, given we are currently looking at an image from minus -199 years ago, and in 14.2 years time we will be looking at the Galaxy at +101.01 years we would see the images updating rapidly.

Yet this would not be fast forward as such - we would be fully aware that we were receiving a barrage of very old images from the Galaxy at high speed due to its approach velocity. It would be like flicking though an album of old photos as you drive to your grand mother's house and just as you arrive you get to the more recent pictures. Time is not accelerated you're looking at the past and flicking through the old images quickly.

We would still conclude that the Galaxy clock advanced by only 2.01 years in a 14.2 year period for us - it would just be that at the start of the journey at Earth time 0.00 we were looking at very old images of the Galaxy from -100 Galaxy years when actually Galaxy time was already simultaneously +99.00;

 

[GhostLoveScore]

If the Galaxy is approaching at 0.99C and the photons at 100%C than they have separated from each other at a speed of 0.01C.

Wait, shouldn't you be adding velocities like this

?

 

[Chestermiller]

If there were two other observers on the train, one at the exact location and time of the right lightning strike and the other at the exact location and time of the left lightning strike, and, if the three observers on the train had synchronized their clocks in advance, and, if the left and right guys recorded their clock readings at the times of the lightning strikes, when they compared notes later, they would find that the right strike occurred first and the left strike occurred afterwards (according to their synchronized clocks).

Chet

 

[Nugatory]

If the Galaxy is approaching at 0.99C and the photons at 100%C than they have separated from each other at a speed of 0.01C

Wait, shouldn't you be adding velocities like this

?

No. That's the relativistic velocity addition formula, which gives you the speed relative to you of something moving at speed $u$ according to someone who is moving at speed $v$ relative to you (read that carefully, several times ). However, in this case the galaxy is moving at speed .99c according to you and the light signal is moving at c according to you, so it is correct to say that according to your clocks and rulers they are separating at a rate of .01c.

We can use the velocity addition to calculate the speed of the light signal according to an observer in the approaching galaxy: Set $u=c$ (the light signal is moving at c relative to us), set ##v=-.99c (as far as galaxy-guy is concerned, we are moving in the opposite direction at .99c), calculate, and you'll find that the light signal is moving at c relative to galaxy guy. (It's a fun exercise to play with the algebra and prove that no matter the relative speed between two observers, they will always find that the speed of a light signal is c relative to them).

 

[GhostLoveScore]

I think I am slowly beginning to understand this. I did not understand a lot of this before, for example, I knew that time slows down by a factor gamma, but I did not take into consideration that length also contracts.

 

[FactChecker]

Everything changes in just the right proportions. "simultaneous" changes, time changes, length changes, things rotate. It all works out.

 

[GhostLoveScore]

If there is an light source and the object moving away from light source at 0.5c, why does the light passes by the object at the speed of light? Is there some thought experiment about this?

 

[PAllen]

If there is an light source and the object moving away from light source at 0.5c, why does the light passes by the object at the speed of light? Is there some thought experiment about this?

Who needs a thought experiment? There are numerous high precision actual experiments on this point.

However, the kind of thinking that led Einstein to guess SR independent of experiment were:

- wave theory suggested that emitter speed has no effect on signal speed. For example, this is true for sound, in the 'air frame'.
- The inability to detect absolute (inertial) motion suggested that you can't detect motion relative to a hypothetical aether (light analog of air).

Putting these together, you are led inexorably to SR. You can also arrive at SR simply assuming isotropy and homogeneity of physical law, along with inability to detect absolute inertial motion. These assumption (nothing about light) lead, by pure logic, to the conclusion that there must be a single invariant speed that is either infinite or finite. Then, any experiment that confirms any prediction of SR (e.g. that muons created in the upper atmosphere reach the ground in large numbers), proves that the invariant speed is finite and that Newtonian physics is wrong (though approximately correct to high accuracy for a range of conditions).

 

[FactChecker]

If there is an light source and the object moving away from light source at 0.5c, why does the light passes by the object at the speed of light? Is there some thought experiment about this?

No. Nobody expected that result and it took a lot of experiments to make people accept it as fact. Even then, people wanted to believe that it was just a trick due to objects being compressed by traveling fast through ether. Only Einstein had enough nerve / genius to assume that the speed of light really was constant and to reach the profound conclusions that it implies. Others had done a lot of the math, but they didn't fully appreciate the consequences. There is no thought experiment to explain it. There are only thought experiments to explain the consequences of it.

 

[GhostLoveScore]

Sorry, I meant why are velocities adding the way they are adding, relativistic way - how could the light sent from moving object still travel at light speed? I know about the equation for calculating that speed, but I'm asking for more intuitive explanation, if there is one.

 

[Stephanus]

Sorry, I meant why are velocities adding the way they are adding

$W = \frac{\frac{U+V}{C}}{1+\frac{UV}{C^2}}$
This is "understandable" for say, V = C

$W = \frac{\frac{U+C}{C}}{1+\frac{UC}{C^2}} = \frac{\frac{U}{C}+\frac{C}{C}}{1+\frac{U}{C}} = \frac{\frac{U}{C}+1}{1+\frac{U}{C}} = 1$, which in SR we always refer to C = 1.
But I still can't make out like the OP question, why the velocity addition formula should be $W = \frac{U+V}{1+UV}$?

how could the light sent from moving object still travel at light speed?

According to whom?
The moving object or the rest observer?
For the moving object, since there is no preferred inertia frame of reference, "the moving object" can consider itself at rest. So the light travels from it travels at ... the speed of light.

For the rest observer viewing the light sent from the moving object?

[..] - wave theory suggested that emitter speed has no effect on signal speed. For example, this is true for sound, in the 'air frame'.
- The inability to detect absolute (inertial) motion suggested that you can't detect motion relative to a hypothetical aether (light analog of air).

Putting these together, you are led inexorably to SR. [..]

The analogy goes further I think. It's like you're at the podium watching a sound wave generated by a race car. The speed of the sound is not [(the speed of the car) + (the speed of the sound)], the speed of the sound generated by the car travels at sound speed, about 340 m/s.
But the analogy stops here.
The speed of the sound wrt car is [(the speed of the sound) - (the speed of the car)]
The speed of light wrt moving object is [(the speed of light) - (the speed of the moving object)]
Is it because the car travels at the same medium and light doesn't travel at some medium?

 

[HallsofIvy]

Your difficulty starts with your first picture! You have labeled "lightning hits engine" and "lightning hits caboose" but you did not say at what time. In the Einstein version, you can take the two lightning flashes as occurring "simultaneously according to a person on the train" or "simultaneously according to a person standing stationary (relative to the ground) off the train". Which do you mean?

 

[HallsofIvy]

Sorry, I meant why are velocities adding the way they are adding, relativistic way - how could the light sent from moving object still travel at light speed? I know about the equation for calculating that speed, but I'm asking for more intuitive explanation, if there is one.

The fact that you are asking this question indicates that you really do not understand the most basic ideas of relativity. The fact that "light sent from moving object still travel at light speed", that the speed is the same in all frames of reference, is a "postulate" derived from experiment, not any calculation, and all other formulas and calculations follow from that.

 

[15characters]

I suspect the OP knows more than it seems as they produced the formula out of nowhere.

This Minowski simulator will remove any confusion

http://www.opensourcephysics.org/items/detail.cfm?ID=11449

 

[harrylin]

If there is an light source and the object moving away from light source at 0.5c, why does the light passes by the object at the speed of light?[..]

The assumption is that the speed of light is c, so that if the object goes at 0.5c then according to your reference system the speed difference is 0.5c.
Alternatively, if we assume that the object is stationary (as is the perspective of a reference system in which the object is in rest), then the speed difference is c.
This was rather well clarified by Nugotory in a recent thread:
https://www.physicsforums.com/threa...trains-speed-relative-to.828679/#post-5205138

Sorry, I meant why are velocities adding the way they are adding, relativistic way - how could the light sent from moving object still travel at light speed? I know about the equation for calculating that speed, but I'm asking for more intuitive explanation, if there is one.

SR is partly based on Maxwell's electrodynamics, which assumes that light propagates as a wave. The light postulate summarizes a key wave characteristic: the speed of a wave is a constant, independent of the motion of the source.

 

[GhostLoveScore]

The fact that you are asking this question indicates that you really do not understand the most basic ideas of relativity. The fact that "light sent from moving object still travel at light speed", that the speed is the same in all frames of reference, is a "postulate" derived from experiment, not any calculation, and all other formulas and calculations follow from that.

Actually, I like asking simple and basic questions even if I understand them because that is the only way I can get some answers. I'm not saying that you people are not helping - you are, thanks for that. But I often get misunderstood if I ask some specific questions and in the end I don't get the answer I was looking for.

 

[HallsofIvy]

You ask questions about things you already know because that way you get answers that you already know?

 

[GhostLoveScore]

As always I'm again misunderstood. In short, no.

 

[15characters]

As always I'm again misunderstood. In short, no.

Did you try the Minowski simulator?

 

[15characters]

The diagram we above shows events now versus events on the moon now (1.25 light seconds away). First diagram shows the timeline for Earth - simultaneous.

Imagine the spaceship is touching down on the moon and the astronaut says "one small step for man one giant leap" his foot hits the ground at exactly 12 noon GMT. The TV image will arrive on Earth in 1.25 seconds. As shown by the light cone on the first diagram.

The second diagram is for an observer passing Earth towards the moon at exactly 12 noon GMT at a speed of 0.99c. For him/her, the touchdown already occurred 8.77 seconds ago on the observer's clock or 1.24 seconds ago on Earth's clocks.

So basically he says we have twice the lag on the radio signal from the moon and that touch down was at 11:59 58.76 seconds GMT not at 12:00 GMT. He also points out that the GMT clocks on the moon are not synchronised with Earth's GMT clocks.

 

[15characters]

Ok this one shows the train and the lightning strikes.

First diagram - you're observer on the train in the middle of the two lightning strikes. In your frame of reference they are simultaneous.

Second diagram - an observer on the platform adjacent to you at precisely the moment you calculate the lightning strike's occurred.

For the observer the back strike occurred 8.77 seconds ago and the front strike will not occur for another 8.77 seconds.

Alternative approach

He will see your front clock reading -1.2375 seconds of what your back clock says.

This is because Velocity x distance = time 0.99 x 1.25 = 1.2375 train seconds

1.2375 train seconds x gamma = 8.772 observer seconds.

So this is why you see the back and front strikes which he considers simultaneous separated by + / - 8.772 seconds respectively.

Reason

The best way to understand it is that the spacetime interval for any two events is invariant; and that between any two events there exists a Proper distance or a Proper Time, which cannot be reduced by any observer.

The interval is like a sort of unique code between any two events, which all observers will agree on whatever they're inertial frame.

(interval = distance - time)

The moment of the lighting strikes was a space like event. This produces a certain type of interval which is the same for all observers whatever their intertial frame - its not possible for anyone even light, to have been at both space like events. There is a fixed distance gap between the events which cannot be eliminated.For the train the events were simultaneous and therefore it can measure the Proper Distance.

No other observer in the Universe can possibly measure a lower Proper Distance unless they share the trains frame of reference. Also, no observer can record Less time between the events than the train because... the train measured 0 time between events.

Therefore any other observer must either add time, or add distance to the trains measurement. And, because of the formula they cannot add one without the other otherwise the interval will increase.

Because the Proper Distance cannot be reduced if any observer wants to use a different reference frame and measurement system the only thing they can do is add extra distance to the train's measurement. This is because the train already has the lowest possible distance and 0 time.

Because the spacetime interval is the same for all observers and the formula is interval = distance - time. If an observer adds time, then they must also add distance. Alternatively, If they add distance then they must also add time.

For the observer on the platform the train's clocks are not synchronised and so the train moves between lightning strikes. The events are therefore separated by a greater distance than for the person on the train

To keep the interval fixed, the observer must add an appropriate amount of extra time between the events to match the extra distance.

interval = train distance plus the observers extra distance - (train's time (zero) + the observer's extra time)

 

[GhostLoveScore]

Did you try the Minowski simulator?

I did not have time today to try it, I just came here to see if there are new posts. I will take a look at it tonight.

 

[15characters]

I did not have time today to try it, I just came here to see if there are new posts. I will take a look at it tonight.

The only way to understand it is to take the time to actually study the subject and actually try to learn something new rather than just asking random questions.

 

[GhostLoveScore]

The only way to understand it is to take the time to actually study the subject and actually try to learn something new rather than just asking random questions.

I'm not asking random questions and I am trying to understand relativity. I am reading lots of different content (e-books, short articles, youtube videos, java applications). I'm sorry if it seems like I have a hat full of questions and every day I pick random one and ask it here.

 

[Nugatory]

I am reading lots of different content (e-books, short articles, youtube videos, java applications)

That may be part of your problem. More is not better here - you will be better off finding one good source and working all the way through it. Physics Forums is an invaluable resource, but it is most effective at helping you over the hard spots in the road when you hit them, not at mapping out the road for you in the first place.

Of course that leaves you with the problem of knowing a good source when you see it. Not everything on the internet is bad, but very little of it has been carefully vetted as an effective teaching tool and coherent presentation of the entire picture, whereas any serious textbook has been... so that's your best bet. You could do worse than Taylor and Wheeler's "Spacetime Physics".

 

[harrylin]

[..] I still can't make out like the OP question, why the velocity addition formula should be $W = \frac{U+V}{1+UV}$?

According to whom?
The moving object or the rest observer?
For the moving object, since there is no preferred inertia frame of reference, "the moving object" can consider itself at rest. So the light travels from it travels at ... the speed of light.

For the rest observer viewing the light sent from the moving object?The analogy goes further I think. It's like you're at the podium watching a sound wave generated by a race car. The speed of the sound is not [(the speed of the car) + (the speed of the sound)], the speed of the sound generated by the car travels at sound speed, about 340 m/s.
But the analogy stops here.
The speed of the sound wrt car is [(the speed of the sound) - (the speed of the car)]
The speed of light wrt moving object is [(the speed of light) - (the speed of the moving object)]
Is it because the car travels at the same medium and light doesn't travel at some medium?

One conceptual model that works for SR is that the car and the light travel (or perhaps one should say "propagate") through the same medium. That's one way to make sense of it.
Another conceptual model that works for SR is that the car and the light describe trajectories through Spacetime, interpreted as a 4D physical "medium".
There were never ending debates on this forum in which people, in vain, tried to disprove either or both interpretations, and a stop of such debates is being enforced here. Remains that you can choose the interpretation you like: whichever fits better with how your brain is wired.
You can find the debates including elaborations of how those models work by means of a search on this forum of such terms as "block universe".