# A question about relativity model

1. Jul 3, 2005

### Ar edhel

I read a recent article in the globe and mail about Einstein theory of motion. it suggested a comparison between a observer's onboard and offboard a moving object, in this case a train. They hinted a very similar model to a ball bouncing on a moving train in my other studies. onboard the train they represent light as traveling a straight perpendicular vertical up and down motion. relative to an observer onboard, the light was required to travel no additional horizontal motion. Then they described how an observer offboard the train would observe the same motion, but also spread across the horizontal distance covered by the motion of the train.

I hope i described the example that was printed well enough. I have a question, mostly to ask if this theory still holds true, and is this still an accurate model... (long question ahead)

I believe this description is very similar to a ball bouncing model we learned about in physics. while a ball bouncing on a train, is not dealing with the nature of light, i saw a few perplexing similarities. As a ball was described, this example deduces using the similar principles. how is it a ball, who's actual speed is not fixed by a constant, be accurately compared against a constant such as light. in order for an observer onboard to bounce a ball up and down (it could be described) as having no additional horizontal motion. and the observer onboard would reflect that it traveled a shorter distance...

But how would this be in any way plausible if the ball was never boosted by the horizontal motion of the train?

Oh and later on in the article it clearly states that motion can neither slow or boost the speed of light. as per Einsteins very words.

Thus i consider it highly improbable that light would take less time to travel, for an observer onboard. because the mere motion alone would not increase the horizontal motion without reducing the amount of motion capable of also exerting vertical... regardless no matter what fraction of time is experienced onboard the train. it is utterly strange for me to conceive that the train would not similarly move forward in the time between up and down intervals, regardless of what smaller accountable time is experienced onboard.

Last edited: Jul 3, 2005
2. Jul 3, 2005

### JesseM

The example you're talking about with a light beam moving up and down is known as a "light clock"--see the explanation here--and it's used to show that in order for both observers to measure the light moving at the same speed c, it must be true that the observer on the tracks sees clocks onboard the train slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$, where v is the velocity of the train as measured by the observer on the tracks.

For a bouncing ball, of course it's no longer true that both observers will measure it travelling at the same speed. If the observer on the train sees the vertical velocity as y and the horizontal velocity as u (u=0 if the ball is bouncing straight up and down), then if the observer on the tracks sees the train going by at velocity v, according to Newtonian physics you'd expect this observer to see the vertical velocity as y and the horizontal velocity as u+v, so using the pythagorean theorem he'd see the total velocity as $$\sqrt{y^2 + (u+v)^2}$$. But in relativity velocities don't add in this simple way--the observer on the tracks will see the vertical velocity as $$y \sqrt{1 - v^2/c^2 }$$ (because if the observer on the train sees the object moving up a distance y in a time 1, the observer on the tracks sees it move up a distance y in a time $$1/\sqrt{1 - v^2/c^2}$$, due to the fact that he sees the train's clocks slowed down by $$\sqrt{1 - v^2/c^2}$$), and he won't see the horizontal velocity as u+v, instead you must use the formula for addition of relativistic velocities (that site is down at the moment so you can also look here for more info), $$(u+v)/(1 + uv/c^2)$$. So, using the pythagorean theorem the observer on the tracks will see the total velocity as $$\sqrt{[y^2 (1 - v^2/c^2)] + [(u+v)^2 / (1 + uv/c^2)^2 ] }$$. You can see that if you apply this formula to y=c and u=0 (a light beam bouncing straight up and down, as seen by an observer on the train), then the observer on the tracks will see the total velocity as $$\sqrt{c^2 (1 - v^2/c^2) + v^2 }$$, or $$\sqrt{ c^2 - v^2 + v^2}$$, which of course reduces to c. On the other hand, if you apply it to a bouncing ball with y equal to a velocity smaller than c, then the two observers will get different answers about its total velocity.

Last edited: Jul 3, 2005
3. Jul 3, 2005

### Ar edhel

Ok it seems straight forward thanks. i may have misunderstood a simple concept about light clocks, i do however have a question, again possibly a long one. (note if you care to read brush up on jesseM's first example) it is undeniable the model you should choose, mine resulting from idle speculation.

i do not see how the motion of the light would reflect in any regard horizontal motion, once emitted it would travel directly upwards. the horizontal motion of the skateboard would not boost the light into following a horizontal path. the reason why it does is because it is chained to a pre guided path within a cylinder. as the cylinder contains horizontal motion, the light within travels an artificially forced increase in horizontal motion, as it may not deviate from its forced route. this is no longer a constant speed, it now contains an artificial method for inducing horizontal motion as well as being required to travel full vertical speed. measuring the distances light traveled is fine... but can you use a constant for light in both equations?

Before you answer. in order to conclude this test the light must also be given the horizontal motion required to remain at the same rate of speed and distance as the observer onboard. by whatever means does it seem likely that light would have traveled horizontal motion by itself? noting that motion may not boost the speed of light.

its hard for me to comprehend, i tend to examine light the instant it is emitted upwards, as a frame... i understand that pausing that moment. light emitted at that point in time would not suddenly change from a purely vertical, towards a vertical and horizontal linear angle. without atleast a force or condition to shape its path horizontal. i wonder if unconditioned to follow a certain path, the light would be required to be sent at an angle to compensate for horizontal motion.

it is not my intent to prove Einstein wrong, in fact i adore his opinions and his life. i have always contained a fond interest of his theories, when i came of the age to understand and fully contemplate them. i however hold back an opinion i believe strongly. it may seem like i blatantly attack peoples opinions, but i do often think best when I'am challenged by other people. I mean no offence to you or anyone that most likely understand's this as fact. i merely challenge other peoples beliefs, cause I'am often not educated enough to understand their explanation... well thanks

4. Jul 3, 2005

### JesseM

If you have a light source which emits light only in a certain direction, like a flashlight or a laser, then it will always emit the light in the same direction in its own rest frame, and if it always emits light vertically in its own rest frame, then from the point of view of an observer who sees the source moving horizontally, the light beam will move diagonally. You shouldn't need relativity to show this, presumably it can be derived as a consequence of Maxwell's laws of electromagnetism which predate relativity.

5. Jul 5, 2005

### Ar edhel

ok im a dullard...

$$\sqrt{1 - v^2/c^2}$$
$$1/\sqrt{1 - v^2/c^2}$$

these formulas represent the increase of distance. to solve or work it out, do they not exactly also measure the amount of vertical motion lost by the exact nature of measuring the total increase as a result of horizontal motion?

deeming that it is easier to discribe that time slows onboard rather then our time increases

Last edited: Jul 5, 2005
6. Jul 5, 2005

### JesseM

The Lorentz contraction formula $$L = L' \sqrt{1 - v^2/c^2}$$ tells you the decrease in the length of a moving object in your frame (L' is the length in the object's rest frame, L is the length measured in your frame), but only along the axis the object is moving. So if something is moving horizontally, its horizontal length will appear to shrink but the vertical length will be unchanged.

7. Jul 5, 2005

### Ar edhel

i think that is simple to understand and straight forward. is that because light of the observer ofboard measures an angle vrs a straight line vertically?

as for the actual length, the theory you put forward simple measures the distances as though light was not forced to compensate for additional horizontal motion, is this the correct answer?

Last edited: Jul 5, 2005
8. Jul 5, 2005

### Ar edhel

sorry to refer to starwars, but if you have a lightsource that emits light slow enough, by this definition you could wave it around like a lightsaber. it would only grow longer. hehe

9. Jul 5, 2005

### JesseM

not sure what you mean by that...just think of a hose spraying water out, it always sprays water straight out from the hose in the rest frame of the hose, but that means someone moving relative to the hose will see the the stream of water moving diagonally. For example, if the hose is pointing down from the ceiling of a train car, an observer on board the train will see the stream of water go straight down to the spot on the floor directly underneath the hose, but an observer on the track sees the train moving, so any water droplet that's part of the stream will have a horizontal velocity as well as a vertical velocity and thus move diagonally in his frame.

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10. Jul 5, 2005

### JesseM

What do you mean by "forced to compensate for the additional horizontal motion"? The length of a moving object can be measured using rulers and clocks--if I have two clocks which are 1 meter apart in my frame and which are also synchronized and at rest in my frame, then if a rod is flying by these clocks and the back end of the rod passes the back clock at the same time the front end passes the front clock (with 'same time' meaning both clocks show the same reading at the moment of these two events), then I say this rod is 1 meter long in my frame. But if it's moving at 0.866c, for example, then in its own rest frame it would be measured to be 2 meters long.

11. Jul 5, 2005

### Ar edhel

the water carrys the inertia of the motion. if light does the same, if you wave a flashlight horizontal is it likely that if it was slowed to a visible state. it would move as as something similar to a lightsaber. in other words, if you imagined you held a slowed lightsource in your hand and slowly moved it left to right. at several different points amongst that motion, would the light all eventually arive at the same location? everything i read supports that light is effected by the inertia of the motion, otherwise the light emitted, early in the experiment, would have gone directly vertical and the experiment would of kept going horizontal without it.

I cant qoute this *i cant even remember whos side of the coin this side lands on* If something has inertia does it not also have speed?\

It would seem that older light would forsake its original path throughout the motion, in order to conduct the test in a different location then when the light was emitted, it would seem that light thoughout the experiment would be required to contain inertia in order go vertical, without being left behind the train.

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12. Jul 5, 2005

### JesseM

Same location where, and in whose frame? Like the same spot on the ceiling of a moving train car where the light source was bolted to the floor? Also, what properties of the "lightsaber" are you thinking of that an ordinary flashlight wouldn't also have?

13. Jul 5, 2005

### Ar edhel

hehe sorry, i am kinda poorly worded in nature. to remain at the same onboard frame, is it likely that as the light moves up and down. horizontal motion is attained, now in order for the light to go up and down. the offboard frame would view that light had to cover enough horizontal motion in order to remain at the same horizontal position as the onboard observer.

i believe it is inertia that causes this because the amount of horizontal distance light travels would almost directly corilate with the speed of the train. so in order to travel the same distance as the scientist onbaord, is it plausible that light would be required to expend a portion of its available vertical speed to remain at the same rate of horizontal movement of the train? given that light may be a constant?

With the increased distance light travels, from the frame of an offboard observer. a stationary clock with a total vertical distance equal to the hypotenuse angle would seem to get the same results...

Last edited: Jul 5, 2005
14. Jul 5, 2005

### JesseM

Yeah, exactly. Did you look at the link to the explanation of the "light clock" thought-experiment in my first post? It's all about a beam of light that's bouncing up and down in one frame, so in another frame it must be moving diagonally.
Its vertical speed $$s_v$$ will have to be less than c, since the total speed of light must be c in all frames, and by the pythagorean theorem the total speed will be $$\sqrt{ {s_v}^2 + {s_h}^2 }$$, where $$s_h$$ is the horizontal speed. Of course, this equation applies for any object that's moving both vertically and horizontally, not just light; it's just that for other objects, their total speed can be different in different frames.
Right, assuming by "same results" you just mean the light will take the same amount of time to go from the bottom mirror to the top mirror.

Last edited: Jul 5, 2005