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# Personal Topic

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Like I said, you talk so much arse!

You think that SR was presented and accepted before it was known that the speed of light is unaffected by the relative velocity of the emitter? Lol!

You said he never said that yet you just admitted it.

you're not thinking properly

What the fcuk is wrong with you?

You said that SR is what showed that the speed of light is constant.

Albert Einstein, in his theory of special relativity, determined that the laws of physics are the same for all non-accelerating observers, and he showed that the speed of light within a vacuum is the same no matter the speed at which an observer travels.

So I pointed out that SR does NOT show that the the speed of light is constant. It describes what happens IF it is constant, something that was obviously already known prior to the formulation of SR. If it wasn't known then there would have been no need for SR.

What's your point? You said "... and he showed that the speed of light within a vacuum is the same no matter the speed at which an observer travels." That's wrong, there would have been no need for SR if it hadn't already been shown that the speed of light is constant in all inertial reference frames. The consistency of the speed of light is a postulate of SR, SR does nothing to validate that postulate other than to show that it actually does work.

Edited by A-wal
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Wow! So does your boat travel close to the speed of light?   :)   It's a speed boat at c.

Apologies for posting this after OceanBreeze has answered so well but I’d already written it and I need clarification on the Doppler Red shift equation. Special relativity means moving clocks run slow

I suspect you'll like this part, Marco:

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Apologies for posting this after OceanBreeze has answered so well but I’d already written it and I need clarification on the Doppler Red shift equation.

Special relativity means moving clocks run slow. Alice’s clock is moving in respect to Bob’s clock on Earth. The time dilation equation is:

∆t0 = ∆t√ 1 – v2/c2

∆t0 = Alice’s spacecraft time interval (in light years for this example)

∆t = Bob’s Earth based time interval in light years = 3/0.6 = 5

v = spacecraft velocity 0.6c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

∆t0 = ∆t√1 – v2/c2     ∆t0 = 5√1 – (0.6)2    ∆ t0 = 5√1 – 0.36         ∆ t0 = 5√0.64

∆ t0 = 5 x 0.8 = 4 years    (0.8 from the equation is Alice’s rate of time due to time dilation, this is also of course ∆t0 /∆t which is 4years/5years in this case = 0.8 for both legs).

Alice would be 4 years older at the end of each leg, while Bob on Earth would be 5 years older at the end of each leg. So Alice would be a total of 8 years and Bob 10years older on Alice’s return to Earth.

I used the Doppler Red shift reference: http://mb-soft.com/public/reltvty1.html

It gave me different equations which I’m not sure of, so I’m confused. My results from using those equations: Doppler Red shift 1 + 0.32 (1.32 octave) on the outward leg, and 1 - 0.32 (0.68 Octave) on the inbound leg,

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I have not said anything to indicate that the age difference "jumps" anywhere.

But if you use instant acceleration just to keep it simple then there will always be an 'instant jump'. Acceleration is always smooth in reality though so there's never an actual jump in practice.

I used the Doppler Red shift reference: http://mb-soft.com/public/reltvty1.html

It gave me different equations which I’m not sure of, so I’m confused. My results from using those equations: Doppler Red shift 1 + 0.32 (1.32 octave) on the outward leg, and 1 - 0.32 (0.68 Octave) on the inbound leg,

In the .6c example being used both Ralfcis and OceanBreeze said that the Doppler shift would be .5 for the outbound journey and 2 for the inbound journey but surely those Doppler shifts are for .5c, not .6c?

I never use Doppler shift because it's non-relativistic and it always cancels out back to zero in the end anyway.

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Apologies for posting this after OceanBreeze has answered so well but I’d already written it and I need clarification on the Doppler Red shift equation.

Special relativity means moving clocks run slow. Alice’s clock is moving in respect to Bob’s clock on Earth. The time dilation equation is:

∆t0 = ∆t√ 1 – v2/c2

∆t0 = Alice’s spacecraft time interval (in light years for this example)

∆t = Bob’s Earth based time interval in light years = 3/0.6 = 5

v = spacecraft velocity 0.6c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

∆t0 = ∆t√1 – v2/c2     ∆t0 = 5√1 – (0.6)2    ∆ t0 = 5√1 – 0.36         ∆ t0 = 5√0.64

∆ t0 = 5 x 0.8 = 4 years    (0.8 from the equation is Alice’s rate of time due to time dilation, this is also of course ∆t0 /∆t which is 4years/5years in this case = 0.8 for both legs).

Alice would be 4 years older at the end of each leg, while Bob on Earth would be 5 years older at the end of each leg. So Alice would be a total of 8 years and Bob 10years older on Alice’s return to Earth.

I used the Doppler Red shift reference: http://mb-soft.com/public/reltvty1.html

It gave me different equations which I’m not sure of, so I’m confused. My results from using those equations: Doppler Red shift 1 + 0.32 (1.32 octave) on the outward leg, and 1 - 0.32 (0.68 Octave) on the inbound leg,

Nothing to apologize for, good post!

If you go back to your link you will see they are calculating the relativistic Doppler redshift in octaves.

One octave is a doubling or a halving in frequency.

They even have a worked problem at 0.6c and the redshift is one octave which is the same as I gave as a doubling in frequency, or 2 in my terminology. I don't know why they are sticking to octaves when talking about light, but that is what they are doing.

In any case, there is no discrepancy.

I hope that clears up the confusion.

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But if you use instant acceleration just to keep it simple then there will always be an 'instant jump'. Acceleration is always smooth in reality though so there's never an actual jump in practice.

In the .6c example being used both Ralfcis and OceanBreeze said that the Doppler shift would be .5 for the outbound journey and 2 for the inbound journey but surely those Doppler shifts are for .5c, not .6c?

I never use Doppler shift because it's non-relativistic and it always cancels out back to zero in the end anyway.

I am not using instant acceleration so there is no sudden jump in age.

Those redshifts are for 0.6c just do the math. I gave the equation

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Those redshifts are for 0.6c just do the math. I gave the equation

I'm not arguing, I just would have thought those were the Doppler shifts for .5c. I don't get why they're for .6c.

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I'm not arguing, I just would have thought those were the Doppler shifts for .5c. I don't get why they're for .6c.

No argument. Can you try plugging 0.6c into the equation I posted?

$Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$

I can give the derivation of that, if you are interested.

Edited by OceanBreeze
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Sadly, I'm looking for experts in the field and the responses I've gotten so far show me I'm in the wrong place. However, if someone has encountered an actual physics forum out there that I haven't been banned from already, I'd be more than happy to check out their recommendations.

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Sadly, I'm looking for experts in the field and the responses I've gotten so far show me I'm in the wrong place. However, if someone has encountered an actual physics forum out there that I haven't been banned from already, I'd be more than happy to check out their recommendations.

You're certainly right that are only a few of us on this forum that have much education in science - and only a subset of those (of whom I am not one) know much about relativity.

I'm tempted to speculate on why you have  - by your own account - been banned from so many places. Usually that is for being either rude or a tiresomely unreformable crank. But I can't acccuse you of that based on your posts here to date.

You could try this place if you have not already:http://www.thescienceforum.com/forum.php

It is not very active but there are a couple of people there that know their SR and GR.

But be warned that moderation there is very brusque.

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No argument. Can you try plugging 0.6c into the equation I posted?

$Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$

I can give the derivation of that, if you are interested.

Oh that's the Doppler shift after applying time dilation, oops. I should have spotted that. If you wanted to know the Doppler shift to apply on top of time dilation then it would be .5 and 2 at .5c.

You can still post the derivation if you like, I should try to get used to seeing it presented formally.

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Sadly, I'm looking for experts in the field and the responses I've gotten so far show me I'm in the wrong place. However, if someone has encountered an actual physics forum out there that I haven't been banned from already, I'd be more than happy to check out their recommendations.

If getting booted from every science/physics forum you have posted on doesn’t give you some second thoughts about what you believe to be true, I don’t know what will.

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Oh that's the Doppler shift after applying time dilation, oops. I should have spotted that. If you wanted to know the Doppler shift to apply on top of time dilation then it would be .5 and 2 at .5c.

You can still post the derivation if you like, I should try to get used to seeing it presented formally.

Yep, you got it. I will post the derivation later as I need to LaTex it.

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You're certainly right that are only a few of us on this forum that have much education in science - and only a subset of those (of whom I am not one) know much about relativity.

I'm tempted to speculate on why you have  - by your own account - been banned from so many places. Usually that is for being either rude or a tiresomely unreformable crank. But I can't acccuse you of that based on your posts here to date.

You could try this place if you have not already:http://www.thescienceforum.com/forum.php

It is not very active but there are a couple of people there that know their SR and GR.

But be warned that moderation there is very brusque.

He will not last long there.

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Ralf;

It seems like you are trying to reinvent the wheel. The aging question was answered long ago, in the 1905 paper where it originated as an example of one clock moving relative to a second clock. Your diagrams at SCF are overly complicated and thus do not effectively communicate your ideas.

So what's new?

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Thanks for the tip, I will try out that forum. If you're really curious of why I've been banned and my threads shut down you can read years of forum interactions on the CR4 general forum and the science philosophy chat forum physics and personal theories. I use ralfcis for every site. I have been continuously wrong over the years and I am able to determine this through math. Two iterations ago I was only off to the third decimal place and could not reconcile the answers I was getting. So I throw out the old and start on a new tack. My current tack has the math working like a well oiled machine and has given me results I did not expect; predictions of testable physics phenomena that relativity cannot make.I have a lot of checking still to do but the math is tedious and I'm very prone to arithmetical mistakes. I used to bother with formulas but now I only deal with STD's which graphically represent the math. People hate them, don't understand them and don't read them.

Like I said I was only on the physicsforums for a day because they are very brusque and they did not like the leading question I asked. I was still confused about the difference between time dilation and age difference and they are not. That's why I want to get back there. I learned from a guy who is the only one I've met who really understands relativity, he wrote a book on it. It took years for him to beat out of me the ideas most people hold on to. The hardest one is the chronic confusion between age difference and reciprocal time dilation and coordinate time. Why would I go back to the way everyone else thinks of them when it took so long to beat it out of me?

Edited by ralfcis
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I promised Awal I would post this derivation, so here it is:

The derivation of   $Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$ comes about from these considerations:

1. The clock on Alice’s spaceship is running slower, as seen by Bob the earth twin, according to the familiar Lorentz transform. It will take $\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ seconds before the next tick of the spaceship’s clock and the next frame of video to be sent (assuming one frame per second for simplicity)
2. The spaceship is also moving away from the earth, in this example, so Bob will not receive that next video frame until the signal, travelling at the speed of light, has travelled back to him over the distance covered by the spaceship between video frames. So, we have the ship moving away at v, and the time between frames is $\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ seconds.

Therefore the distance the spaceship covers between video frames as measured by Bob  is

$\frac{v}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$, as distance is just velocity multiplied by time.

Since the video signal travels at the speed of light c, it covers this distance in time

$\frac{v}{c \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ since time is just distance divided by velocity.

So now we can calculate the total time between Bob seeing one video frame and the next if the frames are generated one second apart according to the clock on Alice’s ship as:

$\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ + $\frac{v}{c \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ =   $Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$

This is the change in frequency which is 0.5 for a halving of the frequency on the way out and 2 or a doubling of the frequency on the way back in. (In octaves, it is a decrease or an increase of one octave)

I hope somebody gets something out of this, but if not my time was not wasted;  I needed to brush up on my LaTex anyway for a paper I am working on regarding marine turbine power systems.

Edited by OceanBreeze
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OceanBreeze i can recognize you're very intelligent. You know more than most people about relativity but you're stuck like I was stuck. If I can find someone in my travels who can answer my test question correctly, I'll come back for you. A second guy backing me up will be far more convincing.

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