Relativity And Simple Algebra

[ralfcis]

Ok I went back to the rear train example and worked out the math there and exposed the Einstein scam for what it is. Please tell me someone on here is capable of understanding high school algebra. 

[marcospolo]

Ok I went back to the rear train example and worked out the math there and exposed the Einstein scam for what it is. Please tell me someone on here is capable of understanding high school algebra. 

You guys love playing around with math and funny diagrams. Too bad it has absolutely nothing to do with reality.

Still nobody answers my question....

I guess that means everyone agrees that its all a big game of fantasy.

[sluggo]

Ralf;

 

So 1 ly was covered in .625 yrs which means v+c = 1.6c which goes against the velocity combo law for relativity.

 

[covered by what?]

[This is a common problem used by anti-relativity protesters, misinterpreting relative speeds as actual object speeds. In the manner of the burger commercial, "where's the object".
Let's do a a simple chase seen. A bank has been robbed, the getaway car A was just spotted 1 mile distant. The police begin their chase. Car A can do 50 mph, while the police car B can do 60. How much time for B to catch up to A?
Typical algebra solution: 60t = 50t+1, t = 1/10 = .1 hr = 6 min.
Wearing your favorite thinking cap, is there anything in this scenario that is moving at 10 mph?

[Returning to the train example:
The problem involves 2 things in motion, a light pulse and a detector at the rear of a train.
The detector moves at .6c, from x = -1 ly to the platform located at x = 0.
The pulse moves at c from x = 0 in the -x direction.
The question is: how much time elapses on the platform clock before the pulse meets the detector.
Typical algebra solution: t+.6t = 1, t = 1/(1+.6) = .625 yr.
There is nothing physical moving at 1.6c, therefore no violation of postulate 2.
These cases are typically classified as 'closing speeds', rates of change of spatial gaps.]

[ralfcis]

Closing speeds are viewed perpendicular to the two things engaged in relative velocity which is a vector. If you go towards a sound wave at 10 mph and the sound wave is going at 750mph, your relative velocity to the sound wave is 760mph. This doesn't mean you have broken the sound barrier. If you go at 10 mph towards a light wave, your relative velocity to the light wave is c not c+10 and this has nothing to do with closing speed. Relative velocity is the starting distance divided by the time it takes for the two parties to meet. In my example, the time to meet is .625 yrs and the starting separation is 1 ly. If the train end had not been moving forward, the time to meet over that distance would have been 1 yr. 

Edited December 21, 2019 by ralfcis

[marcospolo]

Closing speeds are viewed perpendicular to the two things engaged in relative velocity which is a vector. If you go towards a sound wave at 10 mph and the sound wave is going at 750mph, your relative velocity to the sound wave is 760mph. This doesn't mean you have broken the sound barrier. If you go at 10 mph towards a light wave, your relative velocity to the light wave is c not c+10 and this has nothing to do with closing speed. Relative velocity is the starting distance divided by the time it takes for the two parties to meet. In my example, the time to meet is .625 yrs and the starting separation is 1 ly. If the train end had not been moving forward, the time to meet over that distance would have been 1 ly. 

you say two different things here, contradicting yourself.

 

1/   "If you go towards a sound wave at 10 mph and the sound wave is going at 750mph, your relative velocity to the sound wave IS 760mph. This doesn't mean you have broken the sound barrier."

 

 This means that the sound wave speed is NOT measured relative to YOU, it means the sound wave is measured relative to the MEDIUM you and it are moving in!

 

2/  Contradicting yourself in the second statement with:

  " If you go at 10 mph towards a light wave, your relative velocity to the light wave is c NOT  c+10"

 

Which is correct? statement 1 or statement 2?  Is it 1,"wave speed" plus v, or is it "wave speed" alone,  ignoring v totally for no reason at all?

 

Make up your mind.  I wish you guys would pick a team and settle with it!

 

Correct in statement 1, you are only moving at 10 mph, and not breaking the sound barrier. However what speed would you measure the light closing at?  It's 760mph. That is the relative closing effective speed, that's what you get if you measure!

 

But this also must apply to statement  2,  you are only moving at 10,  and light is moving at c (RELATIVE TO THE MEDIUUM< NOT RELATIVE TO YOU!)  then you MUST MEASURE the relative closing speed as c+10, as that is the effective measured speed of light relative to you and is the correct result you will get when you measure.

 

Meanwhile, both you and the light are moving at 10 and c respectively relative to the medium!

 

This is the only sensible math base you need to figure everything out, no need for STD's which are an abortion of geometry.

Edited July 4, 2019 by marcospolo

[ralfcis]

So sluggo, if you're right and there's no problem since it's only closing speed, why would Einstein have gone to so much trouble to arrange the math such that it ensures the train velocity does not add with c to make the relative velocity 1.6c if that number is closing speed as you said. If you check my STD, you'll see using proper simultaneity to sync the clocks does not mask out the train's velocity as Einstein's method does using perspective simultaneity. 

Edited December 21, 2019 by ralfcis

[ralfcis]

Let me try to articulate in simpler terms. Say you have a guy throwing a football at you from 15 ft away and it takes 3 seconds to reach you. The relative velocity between you and the football is 5 ft/s. If you run towards him in 3 sec while he's holding the football, the relative velocity between you and the football is still 5ft/s. But if you run towards a thrown football, you catch it in 1.5 sec and your relative velocity to the football is the original distance, not the distance where you caught it, divided by 1.5 sec = 15/1.5 = 10 ft/sec. If the football had a message written on it, your relative velocity to the message has doubled even though the speed of the football is unchanged.

 

Light is a message. The sun's light takes 8 min to reach us. If the sun goes out, Venus will get that message 2 min before we do. If someone takes off from earth at a serendipitous velocity and time such that he reaches venus just as the light goes out there, he will have received the message before earth does. If relative velocity is original distance/ time you get the message, the relative velocity to the message exceeds c just like in the football example. This is not allowed.

 

Let's change the example to the sun being 1 ly away and the ship leaves earth towards the sun at .6c serendipitously at the exact  instantaneous moment the sun goes out. He has no way of knowing about this accident of proper  simultaneity. If you do the math, the relative velocity is 1.25/1.04167 = 1.2c. But in Einstein's tricky math, his example allows the serendipitous time to be perspectively simultaneous with the sun going out.  It turns out this serendipitous time of take off happens .52 yrs before the sun goes out. The math magically works out by manipulating vt such that c=x/t = 1.5625/1.5625 and x'/t' = .78125/.78125 which proves his assumption that the speed of light is constant from all perspectives (in any example that assumes this and uses his clock sync method to label times on his velocity lines). 

 

It turns out this assumption is not needed to get the right answers as is shown by the method I use. Mine preserves the other absolute must have: that co-located clocks are not subject to perspective time so whether they start or end a spacetime path, they should have no permanent time difference between them. Einstein's choice of start time to artificially preserve the constancy of c from all perspectives using his absurd clock sync method to create a perspective present where all clocks have the same readout despite separation between them is done at the expense of co-located clocks not having the same times on them. I will give him credit for being able to figure all this out to make the math work to support his theory almost. What I don't get is why no one else has figured out his slight of hand.

Edited December 21, 2019 by ralfcis

[sluggo]

Ralf;

 

1. We know the yellow light signal crosses the distance inside the train in 1 yr train time or 1.25 yrs platform time. So in order for c to be constant from all perspectives, the train length from the train's perspective must be 1 ly proper length and the 1.25 ly perspective is really the platform's since 1.25 is the platform time. The light in the train travels the train's proper length in the train's proper time so the length can't be 1.25 and the time 1. Einstein really botched this up because for length contraction to be true, the train's rest length must be 1.25 ly long. This makes c from the train's perspective 1.25/1 = 1.25c which it can't be.

 

 

[in the 1905 paper, Einstein explained length contraction (lc) results from the measurement process, simultaneous for both ends of a rod, i.e. not a physical transformation. The graphic shows the difference it makes.
Left, without lc, there is no reciprocal measurement for P, and the factor is 100/156 = .64 when it should be .80 .
Right, with lc the reciprocal effect is present.]
 

x = x'/Y + vt   and t = t'/Y + vx/c2

 

 

should be x' = g(x-vt) and t' = g(t-vx/cc), with primed frame in motion


x'= 1.25(1.88-.6*3.31) = 0

t' = 1.25(3.13-.6*1.88) = 2.50

in agreement with the graphic 828.

[ralfcis]

Sluggo are you engaged in this discussion or just interested in handing out pre-printed flyers without any point?

Edited July 5, 2019 by ralfcis

[sluggo]

So sluggo, if you're right and there's no problem since it's only closing speed, why would Einstein have gone to so much trouble to arrange the math such that it ensures the train velocity does not add with c to make the relative velocity 1.6c if that number is closing speed as you said. If you check my STD, you'll see using causal simultaneity to sync the clocks does not mask out the train's velocity as Einstein's method does using perspective simultaneity. 

He didn't have to arrange anything. As shown in 838, the c-v, and c+v speeds involve 2 elements in motion, and the solution is equivalent to closing spatial gaps. A gap is a relation of positions. Positions are not regulated by SR. The light spot on the distant wall is not restricted to c. Gaps are not physical elements, thus, in the examples, there is nothing moving faster than c.

[ralfcis]

I read your wrong explanation the first time and your answer just repeats it and is non-sequiter to the point. You are wrong in your interpretation of what closing speed is. It is from a 3rd party perpendicular to the two engaged in relative velocity. Two light beams closing in on each other have a relative velocity of c but  a 3rd party observing them close will see a closing speed of 2c. If you can't understand that then we have nothing further to discuss. We've rarely ever had a back and forth discussion as you rarely answer any questions or follow the discussion from one point to the next. You have no idea what to respond to anything off script. It's like having a discussion with a book.

Edited July 5, 2019 by ralfcis

[marcospolo]

I read your wrong explanation the first time and your answer just repeats it and is non-sequiter to the point. You are wrong in your interpretation of what closing speed is. It is from a 3rd party perpendicular to the two engaged in relative velocity. Two light beams closing in on each other have a relative velocity of c but  a 3rd party observing them close will see a closing speed of 2c. If you can't understand that then we have nothing further to discuss. We've rarely ever had a back and forth discussion as you rarely answer any questions or follow the discussion from one point to the next. You have no idea what to respond to anything off script. It's like having a discussion with a book.

and that 3rd party is in reality, a preferred, absolute stationary frame of reference! 

And EInstein required it in his 1905 paper!

[ralfcis]

The MMX is very much like the train rear example but what's really cool about it is that the train perspective (moving) is physically separate from the platform perspective (stationary) so you can compare the perspectives side by side without having to switch between them.

 

Here's what the MMX would look like in terms of the rear train example. The assumption was light propagated through a medium and if we moved relative to that medium, the resultant relative velocity between us and the light would be higher than light speed. But the results were our relative velocity to the medium did not boost c. It's as though c was a full bucket of water and no matter how much water we added, the bucket wasn't going to get any fuller.

 

But there were effects our velocity had on the light signal such as a change in the light's frequency due to the Doppler effect. It also affected how long the light would have to travel in the moving tube as opposed to the stationary one perpendicular to the light in the moving medium. The travel time for the light in the moving tube should have been much shorter due to the increased velocity of c yet the experiment proved no change in the light speed hence Lorentz had to conclude the moving tube had to have shrunk for this to be true. Einstein concluded that not only did the tube shrink but the time was dilated giving the light more time to cross the shrunken distance. So if the tube shrinks by half but if you give light twice the time to cover that distance you end up with the same light velocity in both tubes.

 

This is a popular misconception in that in any other example of relativity does the length contract in conjunction with time dilation. You use either time dilation or length contraction to solve any of the other examples of relativity. This doesn't even happen in the train example and is not the reason given why the light in the stationary tube meets the light in the moving tube at the same time. The movement of the tube does not shrink the tube, it shortens the distance the light has to travel to the end of the tube. If I'm 200km from Montreal and drive a car halfway it doesn't mean I've length contracted.

 

What's going on in the MMX is my car ride should make me receive the light signal sooner than if I stayed in Ottawa yet the MMX proved both light signals are received simultaneously. Using my method in the train example this is true using proper simultaneity. The light signals start off properly simultaneous and end up properly simultaneous at .833 proper time in both frames. But that time is .625 platform yrs from the platform perspective and 1 train yr from the train perspective. Einstein used the train perspective as the basis of his analysis to get different results that also masked the effects of the train velocity on light speed to magically keep that relative velocity at c. Both interpretations, although wildly different, show how light takes the same time to move over different lengths without affecting c with the velocity of one of those lengths.

 

I have a 3rd interpretation that is much simpler using Yv and Yc which allows normal addition of velocities because the relativistic combo law is baked into the math. I've already touched on this method but will now fully develop and check the new math behind it.

Edited December 21, 2019 by ralfcis

[ralfcis]

So sluggo your argument is Einy didn't purposely make what you define as closing speed look like relative velocity, it just happened to turn out that way. My argument is Einy knew the difference between closing speed and relative velocity and he knew his v+c showed a relative velocity greater than c and he had to hide that fact to keep the rest of his theory from unravelling. 

 

I'm wrong here in that the formula for (c+v) = c/(Y(DSR)) keeps the addition of c+v in check because the sign of v determines the value of DSR. If v and c are approaching each other head on, v or c is negative and DSR is an integer which forces v to be a negative fraction of c according to this formula:

 

v= c/(Y(DSR)) -c. If DSR =2 v=-.6. If DSR = 1/2, v=+.6.

 

If c is trying to catch up with v, DSR is a fraction which forces v to be a positive fraction of c. So the addition of v+c always ends up a fraction of c.

 

In fact relativity never made the claim that plugging c into the velocity combo equation results in the conclusion that adding any velocity to c results in c. In fact plugging  u=c into w=(v+u)/(1+vu/c2) results in v+c = v+c or plugging in w=c results in c=c, never v+c=c.

 

I still believe c is the max speed and information can't travel faster than c but I also believe if you travel towards the light you get the information sooner than someone who doesn't. The starting distance is unchanged but the time has changed so  the concept of relative velocity says the relative velocity to c must have increased but relativity says it could not have. You explain this paradox by saying I'm comparing closing speed to relative velocity. If only this were true then there would be no paradox. But I think Einy's math recognized the paradox and he hid it on purpose.

 

Now why can't you explain your point of view in your own words instead of  having to depend on copying from a book?

Edited December 26, 2019 by ralfcis

[Amplituhedron]

So sluggo your argument is Einy didn't purposely make what you define as closing speed look like relative velocity, it just happened to turn out that way. My argument is Einy knew the difference between closing speed and relative velocity and he knew his v+c showed a relative velocity greater than c and he had to hide that fact to keep the rest of his theory from unravelling. 

 

I still believe c is the max speed and information can't travel faster than c but I also believe if you travel towards the light you get the information sooner than someone who doesn't. The starting distance is unchanged but the time has changed so  the concept of relative velocity says the relative velocity to c must have increased but relativity says it could not have. You explain this paradox by saying I'm comparing closing speed to relative velocity. If only this were true then there would be no paradox. But I think Einy's math recognized the paradox and he hid it on purpose.

 

Now why can't you explain your point of view in your own words instead of  having to depend on copying from a book?

 

You really are a nasty, conceited little creep, aren’t you?

 

Wow! NO ****! So if someone turns on a flashlight and I am running toward that person, and someone behind me just sits still, I will see the light first! Wow!  Just imagine!

 

And this fact violates “Einy’s” theory of relativity HOW, exactly? 

 

You’re a nut!

[ralfcis]

I explained how dumdum. Not my fault if you can't and don't want to understand. You're no different than Marco just at the opposite end of the blind spectrum. At what point did I misrepresent Einy's explanation for the facts of the MMX and the train example.

Edited July 6, 2019 by ralfcis

[ralfcis]

Maybe too many words confuse you, that's common for those with high double digit IQ's. If you reach something in less time your relative velocity is higher. No? Are you stuck at this point? Yet Einy said no relative velocity can be higher than c. So v+c = c always and Einy fudges the math to ensure this but in every example that Greene does there is a term of v+c not equal to c. I suppose your reading and reasoning disabilities extend into comprehending videos as well. The MMX proved your velocity may reach the light sooner from your perspective but your velocity created no difference between light travel time over the distance you made shorter or if you hadn't moved. This is because you are trying to compare different time units for different perspectives. If you use proper time, the time the light travels over the different distances is the same. How it does this has nothing to do with length contraction as you so often vehemently deny without any understanding of math to back up your claim.

Edited December 21, 2019 by ralfcis