# Thread: Theory of Relativity Question

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## Theory of Relativity Question

I guess this is more of a physics question than an astronomy question, but it does involve a spaceship!

I am trying to understand the principals of the theory of relativity as it relates to time. I have seen various explanations using the light clock. The clock consists of a light source and a mirror placed directly above it at some height. A pulse of light is shone vertically from the source, reflected off the mirror and back down to the source. The time this takes is counted as one unit.

The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??

In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?

Sine the light hitting the mirror is a real, visible event, would the astronaut not see it reflect off the mirror at exactly the same moment in time as someone in the spaceship?

When I fire a bullet horizontally from a gun, it falls to the ground in the same amount of time as a bullet dropped vertically from the same height. The amount of extra horizontal distance covered by the fired bullet has nothing to do with the vertical motion. What is the difference between this and the light clock?

Any help would be great!

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even if you stop it at the mirror (and remember, you only see something when it gets absorbed!) the fact remains: The shipboard people will see it have traveled a shorter distance than the stationary observer. And they will both measure it at the same speed...

and that's where the time and legth effects come in.

3. ## Re: Theory of Relativity Question

Originally Posted by thunderchicken
The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??
No. You are thinking logically, which does not work. The important aspect of special relativitiy is light's speed can only be observed to be one speed regardless of any other motion.

You assumed it would take more time for the light to have traveled to the mirror from the viewpoint of the non-spaceship observer. Let's consider first the distance only involved. The distance vertically (as seen by the spaceman) we'll call... d. The distance the light travelled as seen by the observer....d'. The spaceship traveled a certain distance as seen by the observer.....d'2, but this is simply the velocity of the ship times the time the observer saw it move, therefore, it can be stated as .....vt'.

These three distances form a right triangle. d'^2 = (vt')^2 + d^2

But d' = ct' and d = ct. Substitute and solve for t'.

This site is quite helpful.... here

This shows that time is not the same for both observers.

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## Re: Theory of Relativity Question

Originally Posted by thunderchicken
The light clock is placed in a spaceship moving at a very high speed. To an observer in the spaceship the light takes one unit of time to complete its trip. To an observer out side the spaceship, however this is not the case. To him, the light appears to travel in a V. The light is traveling a longer distance, and therefore must take a longer amount of time to complete its trip. Have I understood this correctly??
Yes, this bit is perfect. The outside observer sees a longer amount of time between the sending and return of the photon. As you have described the matter, the light clock is positioned by the pilot to be sending light perpendicular to the direction of motion.

You can calculate the relativistic time dilation with your method. The pilot sees the photons travel for time t. In that time, they travel a distance ct, where c is the speed of light. You call this direction "vertical", while the direction of the ship is "horizontal".

Suppose the ship is seen from outside to be travelling at speed v, horizontally. The photon is now actually moving at an angle, in the V shape you describe, and the outside observer sees it moving for a different amount of time t'. The vertical component is still the same (ct), and the horizontal component is give by the velocity of the ship and the total time of the journey seen from outside, which is vt’.

Pythagorus tells us the photons travelled a total distance sqrt((ct)^2 + (vt’)^2).

Curiously, however, the outside observer still sees the photons travelling at the speed of light. Hence ct’ = sqrt((ct)^2 + (vt’)^2). Divide both sides by ct, and you get:
t’/t = sqrt(1+(v/c)^2))

This is the time dilation factor, and your example so far is perfect, allowing the time dilation to be calculated correctly. t’ is greater than t, so the observer sees the pilot’s clock running slow.

Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?
There is no problem with the light traveling on an angle for one person but not another. Neither case is the “actual” case; each perspective is just as good as the other. You get the same thing in conventional Galilean perspectives.

Sine the light hitting the mirror is a real, visible event, would the astronaut not see it reflect off the mirror at exactly the same moment in time as someone in the spaceship?
No. This is a key point in many of the so-called “paradoxes” of relativity. Not only is time and distance relative; so too is simultaneity. Two events which occur at the same instant for one observer may have a time lag between then for another observer.

For example; from the perspective of the spaceship pilot, the outside observer’s clock is running slow. Each one observers the other clock running slow.

How can this be? The explanation deals with relativity of simultaneity, and is best explained with the spacetime diagrams.

Cheers -- Sylas

5. ## Re: Theory of Relativity Question

Originally Posted by George
No. You are thinking logically, which does not work. The important aspect of special relativitiy is light's speed can only be observed to be one speed regardless of any other motion.
LOL

Originally Posted by thunderchicken
In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?
It is counter intuitive, light doesn't behave like low speed bullets. A photon always travels at c for every inertial observer (one not accelerating). Imagine you fire a photon at c from you and it passes me while I whiz away from you at 0.9c - I will measure the photon as overtaking me at c and that is really bizarre. This is not a result of SR, but an assumption of SR which is backed experimentally and by Maxwell's equations (which is why the assumption was made). Making that counter intuiitive assumption leads to all sorts of wonderful (and counter intuitive) results like time dialation, lorentz contraction and relativity of simultaneity.

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## Re: Theory of Relativity Question

Originally Posted by thunderchicken

In the spaceship’s frame of reference the light is traveling directly vertical, therefore the vertical component of the velocity is C. In the astronaut’s frame of reference the vertical component of the velocity does not seem to be taken into consideration. The speed of light is taken as traveling on an angle, and therefore taking more time to complete the trip. Why is this, since the light is not actually traveling on an angle, it only ‘appears’ to be?
The earth is such a “spacecraft” moving through space at very high speeds relative to other bodies. The television industry has a large number of satellites that are located about 23,500 miles above the surface of the earth. Since the earth-orbit time for the satellites at this distance is about 24 hours, they keep pace with the rotation of the earth and they each hover over a different spot on the earth’s surface since they are rotating or “revolving” with the earth. These are called “geostationary” satellites. It’s like having a satellite at the very top of a 23,500 mile tall TV tower. The TV signals sent up and back down basically go in a fairly straight line.

If these signals were seen from the side, from the point of view of a distant star or planet that has a sideways or lateral motion relative to the earth of, say, 1,000 mps, while the earth and satellite remain in the center of view, then the signals would appear to move not as a “V” shaped line but as an “I” shaped line that travels up and down vertically off and back to the same place on the surface of the earth, while at the same time this line is seen traveling sideways through space along with the earth. It doesn’t matter if we consider the earth to be moving at that speed, or the distant planet to be moving at that speed or the speed to just be relative between the two.

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Doesn't the graviational field also affect time dilation?

I believe that global positioning satellites must use the relativity equations to compensate for a time difference due to the difference in the field strength at the surface of the earth and in orbit. I've been told that if this was not done, you GPS would be off by several kilometers

Is this correct?

Later

Pete

8. GPS satellites must compensate for both gravity and velocity!

9. ## Re: Theory of Relativity Question

Originally Posted by Sam5
If these signals were seen from the side, from the point of view of a distant star or planet that has a sideways or lateral motion relative to the earth of, say, 1,000 mps, while the earth and satellite remain in the center of view, then the signals would appear to move not as a “V” shaped line but as an “I” shaped line that travels up and down vertically off and back to the same place on the surface of the earth, while at the same time this line is seen traveling sideways through space along with the earth.
And within that inertial frame travelling laterally to the earth at 1,000 mps the up-down motion that you described combined with the sideways travelling through space that you described combine to have the light moving a longer distance in a V, at c, relative to that frame, according to relativity, and all the experiments that back it.

The OP's question was about the theory of relativity. I think you should point out that your views are against the mainstream when you attempt to answer basic questions about relativity with something that sounds decidedly like you're disagreeing with the theory. You wouldn't want people to mistakedly think that your take on SR is the convention, would you?

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## Re: Theory of Relativity Question

Originally Posted by worzel
And within that inertial frame travelling laterally to the earth at 1,000 mps the up-down motion that you described combined with the sideways travelling through space that you described combine to have the light moving a longer distance in a V, at c, ....
The light doesn’t travel up and down to and from the satellites in a “V”. It travels up and down in an “I”. It doesn’t matter how fast our earth is moving or our galaxy is moving through space or relative to any other object in space.

11. ## Re: Theory of Relativity Question

Originally Posted by Sam5
The light doesn’t travel up and down to and from the satellites in a “V”. It travels up and down in an “I”. It doesn’t matter how fast our earth is moving or our galaxy is moving through space or relative to any other object in space.
True, in the almost inertial frame where the satellites and the earth are stationary the light travels up and down in an I. But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V. According to relativity light always travels at c in an inertial frame. Therefore, a photon that travels between the earth and a satellite must take longer the faster the intertial frame of the observer is to the earth and its satellites due to the longer path the light must travel in that moving frame. Do you disagree that that is what relativity says?

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## Re: Theory of Relativity Question

Originally Posted by worzel
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.
No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”

13. If you had a tower on a train and the train was moving to the right with a given velocity, and you shot a light beam from the bottom of the tower to the top while the train was moving, a stationary observer would see the light move in a capital lambda (/\). That's one of the basic ways to calculate time dilation.

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Originally Posted by Normandy6644
If you had a tower on a train and the train was moving to the right with a given velocity, and you shot a light beam from the bottom of the tower to the top while the train was moving, a stationary observer would see the light move in a capital lambda (/\). That's one of the basic ways to calculate time dilation.
Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up? Like this:

http://users.powernet.co.uk/bearsoft/LtClk.html

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## Re: Theory of Relativity Question

Originally Posted by Sam5
Originally Posted by worzel
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.
No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.

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## Re: Theory of Relativity Question

Originally Posted by Fortis
Originally Posted by Sam5
Originally Posted by worzel
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.
No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.
That’s like saying you are scanning the sky at night with a searchlight. The distant beam of the light seems to move from star to star, sideways, at millions of times faster than the speed of light because that’s what you see from your frame. But that’s just an illusion.

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## Re: Theory of Relativity Question

Originally Posted by Sam5
Originally Posted by Fortis
Originally Posted by Sam5
Originally Posted by worzel
But in an inertial frame which has a significant velocity relative to the earth and its satellites, the light moves in a V.
No, the light doesn’t move in a “V”. The earth stays under the satellite all the time and the signal moves up and down relative to the satellite and the earth. You don’t count any apparent sideways motion of the light beam as part of “the speed of light.”
I'm picking this one up late, so I may have misunderstood where this one came from, but if you're at rest in an inertial frame that is moving relative to the earth-satellite system, then you must include the "apparent" sideways motion in the speed of light calculation. After all, you are making your measurements in your frame, not anyone elses.
That’s like saying you are scanning the sky at night with a searchlight. The distant beam of the light seems to move from star to star, sideways, at millions of times faster than the speed of light because that’s what you see from your frame. But that’s just an illusion.
Not true. How do you measure the speed of a photon in your reference frame? You determine the distance that it travelled in your frame, and divide it by the time it took (in your frame) to get there.

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## Re: Theory of Relativity Question

Originally Posted by Fortis
How do you measure the speed of a photon in your reference frame?
Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.

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## Re: Theory of Relativity Question

Originally Posted by Sam5
Originally Posted by Fortis
How do you measure the speed of a photon in your reference frame?
Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.
If you couldn't see the earth or the satellite, but just observe the movement of the photons in your frame (this is, of course, a thought experiment), then how would you measure the speed, given that all you know is where and when the photon started on its journey, and where and when it stopped? (All in your rest frame, of course. )

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## Re: Theory of Relativity Question

Originally Posted by Fortis
Originally Posted by Sam5
Originally Posted by Fortis
How do you measure the speed of a photon in your reference frame?
Well I certainly don’t count its speed going sideways even though I know the earth is going sideways in space. Nor would I count the sideways motion of a photon as part of the “speed” of a photon going up to a mirror and back down to the surface of a distant planet.
If you couldn't see the earth or the satellite, but just observe the movement of the photons in your frame (this is, of course, a thought experiment), then how would you measure the speed, given that all you know is where and when the photon started on its journey, and where and when it stopped? (All in your rest frame, of course. )
Doh, umm, hmm, well, I suppose I’d say that the photon must be going at “c” up and down in its system, and I wouldn’t count any sideways motion that might result from an optical illusion. It’s like when I see telephone poles that seem to get smaller and smaller in the distance along the side of a long straight highway. That’s sure what it looks like to me, but I’m reasonably certain they don’t actually get smaller the further away they are from me. I certainly wouldn’t imagine a photon bent over sideways and exiting a laser beam at a 45 degree angle like the one in this animation (the animation at the top of the java box):

Or in this drawing:

http://galileoandeinstein.physics.vi.../srelwhat.html

See the slanted photon in the second drawing down on the right? I don’t know of any laser that emits photons at a 45 degree angle and slanted toward the right, like in that drawing or this one:

http://users.powernet.co.uk/bearsoft/LtClk.html

Such a phenomenon would require an old 19th Century “ether wind” to blow the photon toward the right as it exits the laser.

21. Originally Posted by Sam5
Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up?
It would have to, of course. Think about the laser beam as it leaves the device for the final time, from a frame of reference in which the laser is pointed vertically, but moving laterally. During the time the light beam moves from the bottom to the top, the laser will have itself moved to the side, so the exit will be displaced from where it was at the beginning. Any light beam that was not travelling diagonally would hit the side of the tube. If you think about the beam gradually building up to laser strength by bouncing back and forth between the internal mirrors of the laser, you can see that it will in fact select light travelling at a slant as the light to be reinforced.

Of course, we see this effect in reverse directly all the time, as the aberration of starlight.

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Originally Posted by Grey
Originally Posted by Sam5
Oh... are you one who believes the beam will exit the vertical laser diagonally instead of going straight up?
It would have to, of course. Think about the laser beam as it leaves the device for the final time, from a frame of reference in which the laser is pointed vertically, but moving laterally. During the time the light beam moves from the bottom to the top, the laser will have itself moved to the side, so the exit will be displaced from where it was at the beginning. Any light beam that was not travelling diagonally would hit the side of the tube. If you think about the beam gradually building up to laser strength by bouncing back and forth between the internal mirrors of the laser, you can see that it will in fact select light travelling at a slant as the light to be reinforced.

Of course, we see this effect in reverse directly all the time, as the aberration of starlight.
I think there is something that keeps the laser beam moving straight up and down if the laser is fixed and aimed vertically upward at the surface of the earth, while the earth is moving through space at hundreds and even thousands of miles per second, relative to other galaxies. But I’m not sure what this “something” is. It might be a local inertial effect or some kind of “medium-like” effect with a “medium”, perhaps the earth’s gravity field, traveling through space with the earth.

It would be very interesting to place a laser on a moving vehicle, moving across the surface of the earth and see if the beam would actually hit a mirror fixed directly above the laser that is moving at high speed across the surface of the earth with the laser. I’m not sure if we have anything that moves fast enough to be able to conduct this test. Even the space shuttle, moving at what, about 17,000 mph, might not be fast enough to see if a laser beam aimed at the wall across from it hits a spot on the wall that is directly and exactly across from the laser. That speed is only about 4.7 miles per second, and the walls of the Space Shuttle aren’t very far apart.

I think some of the very old thought experiments need to be updated and modified. I’m not “anti relativity”, but I do think some the relativity concepts should be brought up to date in light of the past hundred years of new discoveries.

23. Another thread successfully hijacked by Sam5! :roll: #-o ](*,)

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Originally Posted by Kaptain K
Another thread successfully hijacked by Sam5!
Excuse me, but this thread was dead, way down on the list, and thunderchicken never came back to comment on the confusing stuff the other guys told him. I can’t help it if I make one post directly to thunderchicken and then 8 guys want to challenge me to a thought experiment duel. What am I supposed to do, not respond to their questions? If I get 8 guys talking to me and asking me questions I’m going to be posting responses to all 8 guys. If you’ve got a complaint, complain to the 8 guys for dragging these discussions out. If the 8 guys hadn’t posted messages directly to me and asked me questions, that one post to thunderchicken would have been the only post I would have made on this thread. If they don’t want me to post any more, then tell them to stop asking me questions.

25. Originally Posted by Sam5
Originally Posted by Kaptain K
Another thread successfully hijacked by Sam5!
Excuse me, but this thread was dead, way down on the list, and thunderchicken never came back to comment on the confusing stuff the other guys told him.
I resent that.

Firstly, I am in no way as eloquent or knowledgable as many of the other posters on BA, but being one who struggled not so long ago to understand exactly the sort of question the OP asked, I feel that sometimes I may have something useful to offer. I also remember posting these sort of questions and appreciating the wide range of answers attempting to describe the same thing from different points of view and trying to deal with my particular misconceptions.

Secondly, all of the answers provided in this thread other than yours are in agreement.

Thirdly, the OP asked a question about the theory of relativity, not about Sam5's inability to understand the most basic of SR expositions. The fact that you never point out that you are against the mainstream when you reply to these sorts of questions with what appears to me be a deliberate attempt to confuse the issue is intellectually dishonest.

I can’t help it if I make one post directly to thunderchicken and then 8 guys want to challenge me to a thought experiment duel. What am I supposed to do, not respond to their questions?
How about starting your own thread in ATM clearly stating what part of relativity you don't agree with. Or at least being honest enough to state that you are ATM when responding to threads like this.

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Hi worzel,

I was telling thunderchicken about TV satellite signals that go up and down to the earth, which is similar to the “light clock” thought experiment that has become part of modern science legend, and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally. That is true, and as far as a can tell it doesn’t refute any of “relativity”, and this is not an ATM theory. Ask any TV satellite engineer.

27. Originally Posted by Sam5
...and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally.
This doesn't make sense to me. Can you show me a drawing of a path that has both vertical and horizontal components, but isn't diagonal?

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Originally Posted by Sam5
Hi worzel,

I was telling thunderchicken about TV satellite signals that go up and down to the earth, which is similar to the “light clock” thought experiment that has become part of modern science legend, and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally. That is true, and as far as a can tell it doesn’t refute any of “relativity”, and this is not an ATM theory. Ask any TV satellite engineer.
This would be your satellite engineer who is at rest in the same rest frame (ignoring for the moment the higher order effects of GR) as the satellite and ground station?

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Originally Posted by Grey
Originally Posted by Sam5
...and I explained to him that the signal is actually moving sideways through space in addition to moving up and down. The signal is not moving diagonally.
This doesn't make sense to me. Can you show me a drawing of a path that has both vertical and horizontal components, but isn't diagonal?
Here’s one that is easier to imagine: A highly directional sound emitter is on the floor of a Concorde, and a directional receiving microphone is directly above it on the ceiling. (The sound is picked up and re-emitted by another highly directional speaker back down to the floor.) The sound signal goes up at 1,100 fps and back down in a straight line to the floor, while the signal is traveling sideways relative to the surface of the earth at 2,200 fps, but we don’t add this sideways speed to the speed of the sound inside the Concorde, and the sound signal travels in a straight line up and down, not a diagonal one.

Edit, one sentence added: (The sound is picked up and re-emitted by another highly directional speaker back down to the floor.)

30. Originally Posted by Sam5
The signal is not moving diagonally.
Yes it is, in the frame of the observer moving relative to the earth and satellite.

Originally Posted by Sam5
Here’s one that is easier to imagine: A highly directional sound emitter is on the floor of a Concorde, and a directional receiving microphone is directly above it on the ceiling. The sound signal goes up at 1,100 fps and back down in a straight line to the floor, while the signal is traveling sideways relative to the surface of the earth at 2,200 fps, but we don’t add this sideways speed to the speed of the sound inside the Concorde, and the sound signal travels in a straight line up and down, not a diagonal one.
Well that would be travelling diagonally relative to a someone moving relative to it. But unlikle light, so would the medium that it is moving in, according to conventional theory. And unlike light, the speed of sound isn't constant for all (intertial) observers, according to conventional theory. So while you might find this easier to imagine, it is of little help for someone trying to understand special relativity.

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