Basic Special Relativity question but the math isn’t working

[ESponge2000]

TL;DR Summary

A photon ball is dribbling vertically up and down on a moving ship, the ship traveling horizontally away from an observer at rest, where it takes one light-second to bounce up to down each way. Since the dribble motion is perpendicular to the direction of motion there should be agreement the up-down vertical is one light second distance , no length contraction perpendicular to the direction of motion

So here’s what’s tripping me

Suppose the person not on ship calculates the dribbling photon vertical relative to the traveling ship, which appears from the stationary person to take 2 seconds to go from up to down but at a diagonal of course , But relative to the moving ship it’s a vertical dribble that takes 1 light second , we then have an easy calc of course

What is 1 light second to earth vertically and 2 light seconds from that point to where the ship is perpendicular to 2 seconds later , and 1 second passes for the ship , the ship moves a distance of sqrt (2^2 - 1^2) = sqrt 3 length in 2 seconds , So v=sqrt(3)/2 times c

Now this seems perfect but the travel frame makes no sense

The photon went up to down in what ship observes was a vertical 1 light-second in a 50% contracted length of sqrt(3)/2 Which means it would observe in that time , a light beam a person not on ship Would find to pass at sqrt (1^2 + 0.75) Why would that make sense?

Ok imagine 2 photon vertical dribbles … one is stationary to the nontraveler , one is veriticle to Lorentz factor 2 ship , they both start at the same point

 

[Ibix]

Have you used the Lorentz transforms in your calculations, or just tried length contraction, time dilation ad prayer? The latter approach won't give sensible results since it neglects the relativity of simultaneity. You need to use the full Lorentz transforms to understand what's going on.

 

[Orodruin]

The photon went up to down in what ship observes was a vertical 1 light-second in a 50% contracted length of sqrt(3)/2 Which means it would observe in that time , a light beam a person not on ship Would find to pass at sqrt (1^2 + 0.75) Why would that make sense?

It doesn’t make sense because you applied length contraction in a situation where it is not applicable. No wonder you get strange results. By definition the light signal (people really should stop talking about photons in a classical theory) just bounces up and down in the ship frame. There is no horizontal motion

 

[ESponge2000]

It doesn’t make sense because you applied length contraction in a situation where it is not applicable. No wonder you get strange results. By definition the light signal (people really should stop talking about photons in a classical theory) just bounces up and down in the ship frame. There is no horizontal motion

Because there’s another dribbling photon in another ship behind the moving one that’s stationary to the non-traveler and suppose that one started dribbling up down right when the traveling ship crossed that point , now that one registers 1 light second every up-down but to the stationary observer, question is what does that photon time elapse look like for the moving ship looking in its rear?

 

[Ibix]

So you've got two identical light clocks in relative motion.

The principle of relativity says that either ship may regard itself as at rest and the other as moving. If one measures the other's clock ticking slowly, what does the second ship measure the first ship's clock to be doing if the principle of relativity is true?

 

[Vanadium 50]

people really should stop talking about photons in a classical theory

photon

So there! Let that be a lesson!

 

[ESponge2000]

That’s right but the math here isn’t working

Ship frame: photon is dribbled down 1 light second vertical

Ship frame : another photon located in the same start place falls 1 light second vertical down but the ship considering itself at rest and everything else moving backwards , observes other photon falls down 1 light second while moving backwards sqrt(3)/2 horizontal … that means sqrt ( 1.75 ) the diagonal distance in light seconds ? That’s what makes so sense

 

[ESponge2000]

Oh I got it ! The time scales between reference frames cannot be reconciled, but for both a round trip of the one at rest on the horizontal = 1 one-way of the diagonal, there’s no reconciling till the clocks are brought to the same point in space. The length contraction math has no use till they are in the same resting frame

If the moving ship stops then the photon clock from the moving ship becomes the diagonal of length 1 on a length contracted sqrt (3/2) and with 0.5 time passage for ship established at when the standing observer saw 1 unit of time pass

 

[Dale]

The length contraction math has no use till they are in the same resting frame

It sounds like you are trying to piece relativity together from bits you have heard about here and there.

You should use the Lorentz transform or the spacetime interval. Both contain all of the relativistic effects in a coherent formula.

 

[ESponge2000]

What i am trying to do is use the twin paradox relationships as a go to for understanding the light clocks in different velocities. And length contraction

Exercise 1: simple slow general relativity
A person dribbles up and down a basketball on a train and a person at the station it passes thru watches it while standing outside the train …. Here we gather they both see the dribble speed fairly at the same rate , But disagree on both the distance and speed of the ball … Because the train horizontal movement is part of the motion of the ball and its moving in zigzags to the person at the station

Example 2: the SR: Then we move to photon light clock where here those “zigzags” are not absorbing the up-down speed plus a moving object’s speed as c is unchanged no matter what object it rides relative to observer so we come to understand this means a disagreement on the dribbling rate
Then we are able to continue that not just time discrepancy occurs with light , but also spatial length distortions which is why 50% of light speed won’t result in aging half as fast, but rather 86% of light speed is needed for this and explained by the Pythagorean math with diagonal traveling light beams …. That is then the foundation for where I come to…. How do those light beams keep light and affect the positioning for each the traveling ship but also the still reference frame and how I got here. And we don’t have to say it’s photon dribbles of a light second if that’s a HUGELY deep up and down …. We can just say 1 light nanosecond per dribble if that’s more spatially sound

 

[Vanadium 50]

I do not know what the "twin paradox relationships" are that you are talking about, but if you are not using the Lorentz transformations of the invariance of interval, you are making a mistake.

 

[ESponge2000]

I do not know what the "twin paradox relationships" are that you are talking about, but if you are not using the Lorentz transformations of the invariance of interval, you are making a mistake.

The Lorentz is the 1/ (1-v^2/c^2)^.5 if that’s what you mean I’m using that, but I’m trying to not take it for granted but tear it apart a bit… like why 1/lorentz is sqrt (1-(v/c)^2)) That there’s a Pythagorean triangle and we have this ….
2 objects are in relative motion horizontally but at rest with each other vertically ….. we can set the direction they are in relative motion as the horizontal assuming they are on the same line of travel such that if one did a 180 their minimum distance would be a collision

So that said , then we can assert that the time perceptions all lie on each perceived vertical up-down , except the vertical up-down on one frame carries the other person’s vertical up-down as their hypotenuse which we can set at 1 unit and then (C^2 - B^2)^.5 where C=1 and B gives a relative time compression factor , = V^2 or the horizontal

 

[Vanadium 50]

No, that is not the Lorentz transformation. See https://en.wikipedia.org/wiki/Lorentz_transformation

You complain the math isn't working. The first step is to use the correct equations.

 

[Dale]

What i am trying to do is use the twin paradox relationships as a go to for understanding the light clocks in different velocities. And length contraction

Exercise 1: simple slow general relativity
A person dribbles up and down a basketball on a train and a person at the station it passes thru watches it while standing outside the train …. Here we gather they both see the dribble speed fairly at the same rate , But disagree on both the distance and speed of the ball … Because the train horizontal movement is part of the motion of the ball and its moving in zigzags to the person at the station

Example 2: the SR: Then we move to photon light clock where here those “zigzags” are not absorbing the up-down speed plus a moving object’s speed as c is unchanged no matter what object it rides relative to observer so we come to understand this means a disagreement on the dribbling rate
Then we are able to continue that not just time discrepancy occurs with light , but also spatial length distortions which is why 50% of light speed won’t result in aging half as fast, but rather 86% of light speed is needed for this and explained by the Pythagorean math with diagonal traveling light beams …. That is then the foundation for where I come to…. How do those light beams keep light and affect the positioning for each the traveling ship but also the still reference frame and how I got here. And we don’t have to say it’s photon dribbles of a light second if that’s a HUGELY deep up and down …. We can just say 1 light nanosecond per dribble if that’s more spatially sound

You can do that, but then you cannot complain that the math doesn’t work out. If you want the math to work then you need to use the math that works. Those are the Lorentz transform or the spacetime interval.

 

[ESponge2000]

You can do that, but then you cannot complain that the math doesn’t work out. If you want the math to work then you need to use the math that works. Those are the Lorentz transform or the spacetime interval.

I see the link now it’s very helpful !!!

 

[ESponge2000]

Actually I still run a problem here
https://wikimedia.org/api/rest_v1/media/math/render/svg/26b4b07b6c0633966abf0b498fa806d067903a8b

I will use this character ŷ for Lorentz factor since I can’t get it on my keyboards

x’ = ŷ (x - vt)

let’s apply this to my light clocks
For the non-moving clock there’s a photon1 born at t=0 And it hits a bottom at t=2 . When t=0 x=0 Y=0 but
When t=2, x=sqrt(3) Y’=Y = 1

Another photon2 also starts at t=0 x=0 y=0 but it ends at t=1 and and x = 0 but y=y’ = 1

So what is the X’ for photon1
= 2 * (sqrt(3) - sqrt(3/2) *2) = 2x0 = 0
Ok this is because Photon1 is moving vertical to the moving ship which is at rest in the prime frame … Ok that works

And the t’ = 2* (2 - sqrt(3)/2 * sqrt 3) =
1 Yes that’s correct
t=2, t’=1

Lets now look at Photon2
X’ = 2 * (0-(-sqrt(3)/2)*1 = sqrt(3)
Yes the other photon moved that direction in the Back
t’ = 2 * (1 - sqrt(3)/2*0) = 2

Oh wow we have it !!!! They both judge each other’s light clocks as the one that’s ticking slow ok it makes perfect sense. They both think they each are the resting frame

 

[Sagittarius A-Star]

I will use this character ŷ for Lorentz factor since I can’t get it on my keyboards

You can write $\gamma$ if you use LaTex for writing formulas. You can find on the bottom left side of the editor window the "LaTex Guide" link. Important is the chapter "Delimiting your LaTeX code".

 

[Dale]

Oh wow we have it !!!! They both judge each other’s light clocks as the one that’s ticking slow ok it makes perfect sense. They both think they each are the resting frame

Yes, you have it.

Length contraction, time dilation, relativity of simultaneity and all of the other relativistic effects are pieces of the puzzle. They all fall out of the Lorentz transform or the spacetime interval as special cases. So in anything but the simplest cases you want to use the general formulas. They will automatically simplify when possible.