# Lorentz Transformations (split from why nothing >c)

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Why nothing can go faster than speed of light.

Yes -
At any given aria nothing can go faster than a speed of light.
However,
Based on relativity - objects can go faster than a speed of light.
Hence, even if an object is not moving at a speed of light, there is a possibility that with regards to other object, it could go faster than a speed of light.

Explanation:

Let assume that A represents an object which is not moving in space.

B will represent an object which moves away from A at 0.6 the speed of light.

C will represent an object which moves away from B at 0.6 the speed of light (at the same vector as AB).

Therefore, the relative speed between A and C is 1.2 the speed of light.

Do you agree with that?

Edited by David Levy

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That is a peculiar seperation velocity. Neither object however is exceeding c.

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Yes -

At any given aria nothing can go faster than a speed of light.

However,

Based on relativity - objects can go faster than a speed of light.

Hence, even if an object is not moving at a speed of light, there is a possibility that with regards to other object, it could go faster than a speed of light.

Explanation:

Let assume that A represents an object which is not moving in space.

B will represent an object which moves away from A at 0.6 the speed of light.

C will represent an object which moves away from B at 0.6 the speed of light (at the same vector as AB).

Therefore, the relative speed between A and C is 1.2 the speed of light.

Do you agree with that?

No. The relative speed between A and C is less than c.

Short, layman's answer - if you'd move at the speed of light (which is impossible) your weight would be infinite.

Edited by koti
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No. The relative speed between A and C is 1.0 the speed of light.

Why?

O.K.

Let look at B.

Do you agree that the relative speed between B and A is 0.6 the speed of light?

Do you agree that the relative speed between B and C is also 0.6 the speed of light?

Hence, do you agree that with regards to B, the relative speed between A and C is 1.2 the speed of light?

If the relative speed between A and C is only 1.0 the speed of light, than how could B see A and C moving at 0.6 the speed of light?

Short, layman's answer - if you'd move at the speed of light (which is impossible) your weight would be infinite.

Yes, that is correct between B to A and B to C.

However, that is not relevant to the speed between A to C.

One more example -

If we look left we can see galaxy D which is moving away from us at almost the speed of light.

If we look rightwe can see galaxy F which is moving away from us at almost the speed of light.

So, what is the relative speed between D to F?

Edited by David Levy
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Why?

O.K.

Let look at B.

Do you agree that the relative speed between B and A is 0.6 the speed of light?

Do you agree that the relative speed between B and C is also 0.6 the speed of light?

Hence, do you agree that with regards to B, the relative speed between A and C is 1.2 the speed of light?

If the relative speed between A and C is only 1.0 the speed of light, than how could B see A and C moving at 0.6 the speed of light?

Yes, that is correct between B to A and B to C.

However, that is not relevant to the speed between A to C.

One more example -

If we look left we can see galaxy D which is moving away from us at almost the speed of light.

If we look rightwe can see galaxy F which is moving away from us at almost the speed of light.

So, what is the relative speed between D to F?

Just less than the speed of light. Always less than the speed of light. A third party observer might measure a gap increasing at greater speed than light could cross - but the relative velocity through space is less than the speed of light

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If we look left we can see galaxy D which is moving away from us at almost the speed of light.

If we look rightwe can see galaxy F which is moving away from us at almost the speed of light.

So, what is the relative speed between D to F?

You need to apply the Lorentz transformation to determine what an observer standing on D would observe. When you do that you will see that he will observe the galaxy moving away at less than c.

Edit: I see that I co-posted with imatfaal, his/her explanation is better because it explains that the relative velocity will also be always less than c.

Edited by koti
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Why?

O.K.

Let look at B.

Do you agree that the relative speed between B and A is 0.6 the speed of light?

Do you agree that the relative speed between B and C is also 0.6 the speed of light?

Hence, do you agree that with regards to B, the relative speed between A and C is 1.2 the speed of light?

If the relative speed between A and C is only 1.0 the speed of light, than how could B see A and C moving at 0.6 the speed of light?

Check post #144. If A and B measure their relative speed as being 0.6c and B and C measure their relative speed as being 0.6c, then A and C will measure their relative speed as being:

$\frac{0.6c+0.6c}{1+ \frac{0.6c(0.6c)}{c^2}} = 0.88235... c$

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Edit: Janus beat me to it, but I typed all this so will post anyway ...

... Hence, do you agree that with regards to B, the relative speed between A and C is 1.2 the speed of light? ...

From the point of view of B, the distance between A and C does grow at 1.2 c; but neither A nor C is actually moving faster than c, according to B.

To get the speed of C, as seen by A, takes the proper velocity addition formula.

It is not simply 0.6 + 0.6 - the Universe doesn't work that way. You'll see the formula here: https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity (edit: and above e.g. by Janus).

In short (i.e. no latex): v3 = ( v1 + v2 ) / ( 1 + ( v1 x v2 ) / ( c x c ) )

As you can see, at low speeds (i.e. our day to day speeds), we get:

v3 = ( v1 + v2 ) / ( 1 + ( v1 x v2 ) / ( c x c ) )

v3 = ( v1 + v2 ) / ( 1 + ( small / big ) )

v3 = ( v1 + v2 ) / ( 1 + almost 0 )

v3 ~ ( v1 + v2 ) / 1

v3 ~ ( v1 + v2 )

So sure, run at 6 km/h through a train that's going at 50 km/h, and someone on the ground watching the train (and you) go past pretty much sees you going at 50 + 6 = 56 km/h

But at speeds approaching c, we get:

v3 = ( v1 + v2 ) / ( 1 + ( v1 x v2 ) / ( c x c ) )

v3 = ( v1 + v2 ) / ( 1 + ( pretty big ) / ( big ) )

v3 = ( v1 + v2 ) / ( 1 + almost 1 )

v3 ~ ( v1 + v2 ) / almost 2

What you see is that as v1 and v2 approach c, they add, but are divided by two. (c + c) / 2 = c

In units of c, and your 0.6 ...

C seen by A (and A seen by C)

= ( 0.6 + 0.6 ) / ( 1 + ( 0.6 x 0.6 ) / ( 1 x 1 ) )

= 1.2 / ( 1 + ( 0.36 / 1 ) )

= 1.2 / ( 1 + 0.36 )

= 1.2 / 1.36

= 0.88

Nobody sees anybody go faster than c.

Edited by pzkpfw
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Check post #144. If A and B measure their relative speed as being 0.6c and B and C measure their relative speed as being 0.6c, then A and C will measure their relative speed as being:

$\frac{0.6c+0.6c}{1+ \frac{0.6c(0.6c)}{c^2}} = 0.88235... c$

O.K.

Let me ask you the following:

If the speed between A and C is 0.88235 c, please prove by mathematics how could it be that the speed between B to A and B to C is 0.6 c.

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O.K.

Let me ask you the following:

If the speed between A and C is 0.88235 c, please prove by mathematics how could it be that the speed between B to A and B to C is 0.6 c.

He gives the equation for adding velocities in the post you quoted. In short, speed doesn't add linearly, it just looks like it does at very low speeds.

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One more question.

If we look left we can see that D galaxy moves almost at the speed of light.

We don't see any further galaxies.

So, there are two options:

1. There are no galaxies beyond D

2. There are other galaxies but they are moving at a speed which is faster than the speed of light and therefore we can't see them.

Which one is correct?

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Expansion of the Universe isn't about speed of Galaxies through space.

Ants walking across the surface of a balloon that's being inflated, may appear to be moving faster than any ant can actually walk.

O.K.

Let me ask you the following:

If the speed between A and C is 0.88235 c, please prove by mathematics how could it be that the speed between B to A and B to C is 0.6 c.

Using the correct velocity addition formula, if A considers C moving at 0.88 c and B moving at 0.6 c, then the speed of C from B can be calculated by rearrangement ...

0.88 = (Vbc + 0.6) / (1 + (( 0.6 + Vbc ) / 1))

0.88 = (Vbc + 0.6) / (1 + 0.6 Vbc)

0.88 ( 1 + 0.6 Vbc) = 0.6 + Vbc

0.88 + 0.528 Vbc = 0.6 + Vbc

0.28 = 0.472 Vbc

Vbc = 0.59322

... which is (given all the rounding) basically that 0.6 c we're expecting.

Edited by pzkpfw
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Expansion of the Universe isn't about speed of Galaxies through space.

Ants walking across the surface of a balloon that's being inflated, may appear to be moving faster than any ant can actually walk.

Great analogy.

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Expansion of the Universe isn't about speed of Galaxies through space.

Ants walking across the surface of a balloon that's being inflated, may appear to be moving faster than any ant can actually walk.

Well, we are living in one Universe.

The expansion is an important section of that universe.

Therefore, it must be included in the formula of speed.

We can't just use a formula which doesn't take in account the expansion effect.

Do you agree that there are galaxies beyond D?

If so, why we can't see them?

Why we can't assume that –

Due to the expansion some galaxies are moving at the speed which is higher than c and therefore we can't see them?

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One more question.

If we look left we can see that D galaxy moves almost at the speed of light.

We don't see any further galaxies.

So, there are two options:

1. There are no galaxies beyond D

2. There are other galaxies but they are moving at a speed which is faster than the speed of light and therefore we can't see them.

Which one is correct?

Since you were rude enough not to bother to read my last post, I am not well inclined to answer your question.

However your analysis in incorrect, there are more than 2 options.

I suggest your understanding and/or description of the scenario you present is incomplete as the reason.

Provide a proper and complete description of the situation vis-a-vis object D and you will have your answer.

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Well, we are living in one Universe.

The expansion is an important section of that universe.

Therefore, it must be included in the formula of speed.

We can't just use a formula which doesn't take in account the expansion effect.

Do you agree that there are galaxies beyond D?

If so, why we can't see them?

Why we can't assume that –

Due to the expansion some galaxies are moving at the speed which is higher than c and therefore we can't see them?

Because it would be wrong. The space is expanding in between us and them - they are not moving through space at a relative speed greater than c

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Because it would be wrong. The space is expanding in between us and them - they are not moving through space at a relative speed greater than c

O.K.

The space is expanding in between us and them.

Hence, there are two options:

1. The expansion increases the relative speed (above c) between us and them and therefore we can't see them.

2. The expansion doesn't increase the relative speed between us and them and therefore we must see them.

False dichotomy.

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Using the correct velocity addition formula, if A considers C moving at 0.88 c and B moving at 0.6 c, then the speed of C from B can be calculated by rearrangement ...

0.88 = (Vbc + 0.6) / (1 + (( 0.6 + Vbc ) / 1))

0.88 = (Vbc + 0.6) / (1 + 0.6 Vbc)

0.88 ( 1 + 0.6 Vbc) = 0.6 + Vbc

0.88 + 0.528 Vbc = 0.6 + Vbc

0.28 = 0.472 Vbc

Vbc = 0.59322

... which is (given all the rounding) basically that 0.6 c we're expecting.

O.k.

The mathematics is clear.

However, I still don't understand the logic.

I would like to verify the correlation with the following formula:

S = V * T

S = Distance

V = Velocity

T = Time.

So, let's look at the following example:

The relative speed between B to A and B to C is 0.6c.

Therefore, after t1 the distance between B to A and B to C should be increased by:

S (ba) = S (bc) = S1 = 0.6 c * t1.

As BA vector is on the same line as BC, than by definition the distance between C to A should be increased by:

S (ca) = S (ba) + S (bc) = 2 * S1 = 1.2 c * t1.

It took t1 (time) for C and A to increase their distance by S (ca)

Therefore - during t1 their relative speed is:

V (ca) = S (ca) / t1 = 1.2c

Simple and clear.

Do you see any error with that calculation?

However, based on your mathematics - the relative speed between A to C is:

V (ca) = 0.88 c.

Therefore, after t1 the increased distance between A and C is:

S (ca) = 0.88 c * t1.

That can't represent the real increased distance of

S (ca) = 1.2 c * t1.

Can you please explain this enigma?

Edited by David Levy
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Can you please explain this enigma?

It's the same thing again and again. You can't ask the same question 4 times and expect a different answer the 4th time.

Once again, you're using the view from B of A and C, and applying it to the view from A of C. That can't work the way you want it to. Whether you simply look at speed, or dress it up with distance traveled, you're ignoring that thing called relativity.

Speed, time and distance don't simply add the way you will think if viewed from a naive pre-relativistic way.

Edit: one specific additional point raised in your post is important here ... when you write (my underline) "... That can't represent the real increased distance of ..." what makes you think any one of those distances is any more "real" than another? Why does the increase in distance between A and C as seen by B have to be "real" and why do you demand that the same thing is seen by A of C, or by C of A? This shows how you are ignoring relativity in your thinking. You are looking for absolutes.

Edited by pzkpfw
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Thanks for you support

It's the same thing again and again. You can't ask the same question 4 times and expect a different answer the 4th time.

Sorry, but I ask the same question as I do not get a direct answer.

Once again, you're using the view from B of A and C, and applying it to the view from A of C. That can't work the way you want it to. Whether you simply look at speed, or dress it up with distance traveled, you're ignoring that thing called relativity.

This shows how you are ignoring relativity in your thinking. You are looking for absolutes.

That is correct. I focus on absolutes

1. Do you agree that with regards to B point of view, the relative speed between A and C is 1.2c? Yes or no

2. If so, do you agree that based on B point of view and t1 (time frame) the distance S (ac) between A to C had been increased by 1.2c * t1? Yes or no

If the answers to the above are - yes, let's move to relativity:

Speed, time and distance don't simply add the way you will think if viewed from a naive pre-relativistic way.

Edit: one specific additional point raised in your post is important here ... when you write (my underline) "... That can't represent the real increased distance of ..." what makes you think any one of those distances is any more "real" than another? Why does the increase in distance between A and C as seen by B have to be "real" and why do you demand that the same thing is seen by A of C, or by C of A? This shows how you are ignoring relativity in your thinking. You are looking for absolutes.

O.K.

So based on relativity, the view from C isn't the same as the view from B.

With regards to - Distance.

If we look from B point of view, the speed V (ac) between A to C is 1.2c and therefore the distance had been increased by S (ac) = 1.2c * t1.

If we look from C point of view, the speed V (ac) between A to C is 0.88c and therefore the distance had been increased by S (ac) = 0.88c * t1.

I had the impression that distance by definition must be absolute.

So, which one is the correct answer?

I still don't understand how could it be that due to relativity both are correct.

Edited by David Levy
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....I still don't understand how could it be that due to relativity both are correct.

Why do you spend all this time asking questions just to partly ignore the answers? You do it in thread after thread; ask a question, pick part of the answer and ask a follow-up/alternative questions which makes it quite clear you have not invested the time in actually engaging with the perfect answer already given.

Times, distances, and speeds do not behave in a way that we naive bald apes find natural when relative speeds get to a noticeable proportion of c. We have an entire ancestry who learned how to "get" speed, distance and time in an instinctive manner - and that instinct lets us down when the speeds get high. You cannot add velocity, length is not absolute, time is not absolute, simultaneity is a local construction - it is odd.

Please please please - re-read the answers. You cannot just "get" this - it requires study, hard work, and falling back to experimental data (invariance of the speed of light) to understand

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O.K.

However,

Would you kindly advice if both are correct:

If we look from B point of view, the speed V (ac) between A to C is 1.2c and therefore the distance had been increased by S (ac) = 1.2c * t1.

If we look from C point of view, the speed V (ac) between A to C is 0.88c and therefore the distance had been increased by S (ac) = 0.88c * t1.

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Not sure if this is relevant, but if two electrons are speeding toward each other, they are being repulsed by the EMF. Once they get to 10^-31 cm apart gravity takes over and becomes the dominant force.

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O.k.

The mathematics is clear.

However, I still don't understand the logic.

I would like to verify the correlation with the following formula:

S = V * T

S = Distance

V = Velocity

T = Time.

So, let's look at the following example:

The relative speed between B to A and B to C is 0.6c.

Therefore, after t1 the distance between B to A and B to C should be increased by:

S (ba) = S (bc) = S1 = 0.6 c * t1.

As BA vector is on the same line as BC, than by definition the distance between C to A should be increased by:

S (ca) = S (ba) + S (bc) = 2 * S1 = 1.2 c * t1.

It took t1 (time) for C and A to increase their distance by S (ca)

Therefore - during t1 their relative speed is:

V (ca) = S (ca) / t1 = 1.2c

Simple and clear.

Do you see any error with that calculation?

However, based on your mathematics - the relative speed between A to C is:

V (ca) = 0.88 c.

Therefore, after t1 the increased distance between A and C is:

S (ca) = 0.88 c * t1.

That can't represent the real increased distance of

S (ca) = 1.2 c * t1.

Can you please explain this enigma?

We'll use a example that shows how, time dilation, length contraction and the relativity of simultaneity combine to the give the velocity addition result.

A is an observer with a clock.

B is an observer with his own clock passing A an at the rear of a rocket moving at 0.6c relative to A, by B's measurements, the rocket is 0.6 light seconds long and there is a clock at the front which is synced to his own.

C is a projectile fired by B at 0.6c relative to himself towards the front of the rocket. He fires the projectile at the moment he is passing A.

Thus at the start, A,B, and C are co-located.

something like this.

A

B______|

C

The vertical line represents the front of the rocket.

If we add the clocks it would look like this according to B at the start.

A(0)

B(0)____|(0)

C

After C reaches the front of the rocket things would be like this:

A(0.48)

...........B(1)____|(1)

.........................C

B's clock will have advanced 1 sec while C traveled from B to the front of the rocket, as will the clock at the front of the rocket, so C arrives at the front clock when it reads 1. A and B will have separated by 0.6 light sec, and because A has a relative motion with respect to B, B measures A"s clock as running slow by a factor of 0.8.

Now we consider things according to A, at the start things look like this:

A(0)

B(0)___|(-0.36)

C

Due to length contraction, the rocket is only 0.48 light sec long.

Also, due to the relativity of simultaneity, the clock at the front of the rocket is offset to read 0.36 sec earlier than B's clock.

When reaches the front of the ship, things are like this:

A(1.7)

.................B(1.36)_|(1)

..............................C

C still reaches the front of the rocket when the clock there reads 1 sec just like it did according to B.

In this time both B's and the front of the ship's clocks will have advanced by 1.36 sec. (the clock at the front has to go from -0.36 to 1, and even though they are offset from each other, the two clocks on the rocket still tick at the same rate.)

Since the rocket's clocks will be time dilated by 0.8 according to A, A's clock will have advanced by 1.7 sec.

In that 1.7 secs, at 0.6c, the rocket will have moved 1.02 light secs and C, at the front of the ship will be 1.02+.48= 1.5 light secs from A.

Thus C, which started next to A, traveled 1.5 light sec in 1.7 sec, or at a speed of 0.88235...c relative to A as measured by A.

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