Relativity and simple algebra II

Discussion in 'Alternative Theories' started by ralfcis, Feb 6, 2021.

  1. ralfcis Registered Senior Member

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    Here's the Minkowski diagram translation of the Loedel diagram.

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    Lots to unpack here.
     
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  3. ralfcis Registered Senior Member

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    So a prime rule of relativity is the principle of relativity. This carries over into how relative velocity is depicted in spacetime diagrams, all the physical parameters should be the same no matter how it's drawn. But I can see at least 2 disturbing differences between how the Loedel diagram differs from the Minkowski depiction:

    1. The distance separation between Bob and Alice should be the same. At t'=2, it's 1.5 ly in the Md but only 1.414 ly in the Ld. It's because the stationary frames are different, one is the Earth and the other is a blank empty space. The time units of the blank space are different from Earth's time units so the distance units will also differ. The Cartesian grid is square and it's a different unit size in the two depictions.

    2. The light signals are not the same length even though they represent the same velocity of c. The pink light signal travels twice as far in twice the time (as the yellow) chasing Alice speeding away from it. If I'm going to represent both perspectives on a single spacetime diagram, I'm going to have to compensate for this apparent imbalance in symmetry when depicted in a single spacetime diagram.

    This actually has huge implications. Imagine a football thrown at 10m/s relative to the field. The speed and direction of the catcher determine the relative velocity of the catcher to the football but the football relative to the field remains at 10m/s. Normally the relative velocity of catcher to football would decrease if he's running away and increase if he's going toward it. If the football was light, his movement towards the light can't have a relative velocity to c greater than c. There's no danger of that happening if he's moving away yet his relative velocity to c is still c as shown in the velocity combo formula. The reason it doesn't is because the field is a vacuum and his movement has no relative velocity to the vacuum in any direction. However, in the spacetime diagram, Bob's pink light to Alice chases Alice. It takes that light more time over greater distance to catch up which is the definition of decreased relative velocity except Alice can have no relative velocity to light's medium otherwise her relative velocity to c in the opposite direction would exceed c.

    If the field was water, the ball could only travel 7.5m/s through that medium which would allow the motion of the catcher relative to the medium to affect his relative velocity to the ball.

    I will show how this works mathematically in the next post. The light signals are the same velocity but not of equal duration even though they are sent and received at the same proper time by each participant.
     
    Last edited: Feb 8, 2021
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  5. James R Just this guy, you know? Staff Member

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    ralfis:

    That seems like a negative for your theory, right there. All of special relativity can be derived in a chapter or two of a standard undergraduate textbook. That seems much simpler than however you did it.

    I should also ask: does your theory disagree with special relativity, in terms of any testable predictions? Or is it just a reformulation of special relativity?

    Okay. That's interesting. No time dilation in your version.

    Let's take a standard example, then. Observer A remains stationary. Observer B travels past A at a speed of (3/5)c, heading for a destination that is a distance x away from A, as measured by Observer A.

    Observer A calculates that for B to travel from A to the destination, the time taken (measured on A's clocks) will be $t=\frac{5}{3}\frac{x}{c}$.

    Can you please show me how you would calculate how long it will take B to travel from A's location to the destination, according to B's clocks, using your equations? Is your answer the same as the answer from special relativity?

    If the times for A and B are different, make sure you show how relativity of simultaneity accounts for the difference.

    Thanks.

    Never mind, for now.

    What's the point of having the formula, then, if it's not useful for anything?

    What do you mean when you say "c has 2 velocity components"? c is a constant, isn't it? The speed of light in vacuum?

    How about an example? What if $v=\frac{3}{5}c$ and $v_t=\frac{4}{5}c$? Then what does your "enhanced" velocity addition formula have to say about that?
     
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  7. James R Just this guy, you know? Staff Member

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    ralfis:

    In your replies, can you please use [quote][/quote] tags to distinguish your words from the words of other posts you are quoting? It is very difficult to sort out whose words are whose if you just write it all out the same way.
     
  8. ralfcis Registered Senior Member

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    421
    I answered these questions twice already. My math is much simpler, you haven't seen the full math yet. It does not disagree with the experimental results of relativity as I've said twice already

    . It is a math exercise.

    Never said that. I said there's no time slowing. I emphasize use of relativity of simultaneity.
    I've already shown your example in my spacetime diagram.
    I already showed the derivation in my next post where I recopied your original 2 questions.
    I already explained the usefulness.
    Fully explained. You're just not reading or understanding what I write

    \( c^2 = v^2 +v_t^2\) so c^2=3/5^2 +4/5^2 no?

    You're just not getting enough of this to point out anything so maybe wait until there's more to read. Can you read spacetime diagrams? If not, there's nothing here for you. This is a math exercise. In the future, I'm going to answer any questions that have not been previously answered because there's a lot of math coming and I'm busy.
     
  9. James R Just this guy, you know? Staff Member

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    Sure, but that's not your "enhanced" velocity addition formula. To be clear, I'm talking about this one:

    \(c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2\)

    What are $v$, $u$, $v_t$ and $u_t$?

    I asked you to give a few examples of the application of that formula.
    Yes.

    I'll take a look at the stuff you've posted.
     
  10. James R Just this guy, you know? Staff Member

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    So $v_t$ is a dimensionless number.

    But that doesn't make any sense, since you wrote:

    $c^2=v_t^2+v^2$.

    That equation is dimensionally incorrect if $v_t$ has different units from $v$ and $c$. It doesn't matter how it is derived (or if it is an axiom). It is just wrong.

    Do you understand that the units of all terms in any physical equation must match?

    I have no idea what that means. Are you just saying that no observer sees clocks in his own frame as ticking at anything other than the "normal rate"? That would just be the same as saying that all inertial frames are equivalent, approximately. Is that what you're saying, or are you saying something different?

    Not in relativity. In your version of relativity. Do you have a name for your theory, by the way?

    I don't know what you mean by an "imbalance of relative velocities". How do we "balance" relative velocities? By "balance" are you just saying that two relative velocities are different from one another, or are you saying something else?

    Unless you use precise language, this is going to be a really difficult discussion. As you can see, I'm already struggling to understand your meaning, in many cases.

    You seem to be saying that a "participant" "initiating a change" can cause a "permanent slowing of time". So my questions are:
    • What is a "participant"?
    • What quantity is being changed?
    • What kind of change causes a permanent slowing of time?
    • Does that permanent slowing of time apply to all "participants" everywhere, or only to one of the "participants"?
    • Can time ever speed up again, once it has been permanently slowed down? (The word "permanent" suggests it can't.)
    • Does that mean that the entire universe is constantly slowing down, every time any "participant" makes the relevant change?
    • Why does the permanent slowing happen only when the participant "initiates" a change, and not at any other time during or after the change?
    • Please define "Doppler Shift Ratio" for me. What it is a ratio of?
    • If it is only "partly" an "apparent rate", what is the other part?
    • What kind of "rate" are you talking about?
    • Who is that "rate" apparent to (partly)?
    • What do you mean by "rate of info delay"?
    Is $v_t$ relative, then?

    Could you please give the equation for "your observed time rate"?
    ---

    So, to summarise the above, the three "types of rates through time" are, according to you:
    1. a "permanent slowing of time";
    2. the "Doppler shift ratio"; and
    3. $v_t$.
    At this point, I don't understand how either a "permanent slowing of time", or a "Doppler shift ratio" can be a "rate". Moreover, I don't see how $v_t$, which appears to be a dimensionless quantity, can be a "rate through time".
     
    Last edited: Feb 9, 2021
  11. James R Just this guy, you know? Staff Member

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    Please correct me if I'm wrong here.

    $v$ is the velocity of "the participant".

    If $c^2=v^2+v_t$, which you say is your "main equation", then the following algebra should apply:

    $c^2 = \left(\frac{3}{5}c\right)^2 + v_t^2$
    $\frac{16}{25}c^2 = v_t^2$
    $v_t = \frac{4}{5}c$.

    But you have, instead, $v_t = \frac{4}{5}$, without the "c" factor.

    Something isn't right. Did you make a mistake about the units of $v_t$? Or is your "main equation" incorrect?
    ----

    Moving on, you say "DSR=.5c". Let's see. Looking through your list of formula, I find this one:

    $DSR^2= \frac{(c-v)}{(c+v)}$.

    So, plugging in $v=\frac{3}{5}c$, we find:

    $DSR^2=\frac{(c-\frac{3}{5}c)}{(c+\frac{3}{5}c)} = \frac{\frac{2}{5}c}{\frac{8}{5}c}=\frac{1}{4}$
    $DSR=\frac{1}{2}$.

    But you have DSR=.5c, which has an extra factor of "c" in it.

    Something isn't right, again. Did you make a mistake in the formula for DSR, or is your equation for the DSR, as written, incorrect?
    ----

    Lastly, you say "if the participant was to turn around after 3 ly, she would age 2 yrs less when the two re-unite at Earth (i.e. the twin paradox.)".

    I don't understand why we need $v_t$ or DSR to calculate this. You haven't said.

    But let's check. We're talking about a round-trip of 6 ly, at a speed of (3/5)c, which would take 10 years according to a "stationary" observer.
    Now, if we were using special relativity, the "participant's" clock (i.e. the moving observer's clock) would be ticking slowly according to the "stationary" clock, by a factor of

    $\gamma = \frac{1}{\sqrt(1-(v/c)^2}=\frac{5}{4}$,

    which means the "moving" clock would record an elapsed time of 8 years. In the "twin paradox" scenario, the "moving" observer will have aged 2 years less than the "stationary" observer when they are re-united, so your result matches the result from special relativity, even though I'm not sure how you calculated it.

    Of course, if we're using special relativity, we have to assume that periods of acceleration and deceleration are instantaneous in this scenario. Is that also the case for your theory?

    According to special relativity, nobody in this scenario suffers any "permanent slowing of time". Once the "twins" are reunited, their clocks once again tick happily in synchrony with one another. During the journey, the "stationary" twin observes an apparent slowing of time on the "moving" spaceship, while the "moving" win observes slowing of the "stationary" clocks at all times except during the times when he/she is accelerating, at which times he/she observes an apparent speeding up of the "stationary" clocks.

    There are no Doppler shifts in this scenario, since no messages are being sent from one observer to the other.

    There's no $v_t$ in special relativity, and it seems we don't need it for anything.
     
  12. James R Just this guy, you know? Staff Member

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    Or maybe it's the other way around. It doesn't really matter, either way, I think.

    I don't think it's very helpful to think of a vacuum as being like a material medium (like water, for example). A vacuum is empty space.

    That's a bit muddled. Light certainly has a velocity relative to the bottle. If you rode along on the bottle, you'd measure the speed of light to be c, relative to the bottle. The problem is that you can't ride along on a beam of light, so it doesn't make much sense to say the bottle has a velocity of c relative to the light. I guess if you wanted to strain things, you might say that everything has velocity of c relative to light. That would be better than saying everything has "no velocity" relative to light. But perhaps we're just saying the same thing in different ways?

    You'd need to be careful to use the correct relativistic velocity additional formula, or you'd get the wrong answer.

    I don't really understand this. For example, on a Minkowski diagram, a constant relativity velocity in the frame of the diagram can be depicted as a straight line on the graph.

    Adding a third observer to a scenario with two observers does not make the situation unphysical.

    There's no necessity to add Earth into the picture, if you're only interested in observations about the two ships, made in the reference frame of either of the ships.

    Okay. For some reason you want to use a reference frame in which the Earth is moving. I don't know what you gain from doing that, exactly. Am I missing something?

    Wouldn't that be zero velocity? If there are two velocities $\pm v$, then the "average" of the two is zero, not $v/2$.

    How long have you been at this? You previously said 10 years or something, didn't you? It beggars belief that in 10 years you haven't found anybody who is familiar with Minkowski diagrams.

    You've come to the right place to get some useful feedback, then, if sciforums is the first place you've found that has some members who are moderately competent with special relativity. Hopefully we can connect you to some real relativity.
     
    Last edited: Feb 9, 2021
  13. James R Just this guy, you know? Staff Member

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    39,287
    I don't understand what you're saying. When should the distance between Bob and Alice be "the same"? The same as what other distance (i.e. what two distances are you comparing)? According to which observer? At what time? Using whose clock and whose rulers?

    I also don't know what your abbreviations "Md" and "Ld" mean. Oh wait, do you mean "Minkowski diagram" and "Loedel diagram"? Please don't use abbreviations until you have defined them. You leave your readers to guess.

    Are you referring to the distance between Bob and Alice, indicated at t'=2 on both diagrams? Obviously those distances should be different, because they are drawn for two different reference frames. Your "Loedel" diagram looks at the distance measured by an observer travelling at a speed of 0.33c, relative to both Bob and Alice, whereas your "Minkowski" diagram looks at the distance measured by Bob/Earth between Bob and Alice. Different observers use different rulers in special relativity. Recall length contraction.

    Yes, you're right. It's a consequence of the Lorentz transformation between the two different observer frames. Both time dilation and length contraction effects are seen.

    Not a problem. If the two signals are both travelling at c, then one will travel twice as far in twice the time.

    You shouldn't expect the times to be the same in two different reference frames. If you did, you'd be forgetting time dilation.

    You're already having to compensate, by having to mark off different t and t' coordinates on the various worldlines you've drawn.

    As long as the light travels at c in all frames of reference, there's no problem in terms of relativity.

    Don't forget, when you're adding velocities, to use the correct velocity addition formula from relativity, by the way.
    But water isn't the same as vacuum. There's no requirement that a football travel at the same speed in all frames in water (or even in a vacuum for that matter). There's also no requirement that light should travel at the same speed in water in all frames.

    I'm not seeing the problem you're trying to point out. Maybe when you present your calculations, we'll see a problem.

    I thought, though, that your version of relativity makes all the same predictions as the standard theory of special relativity.

    Now, it seems like you're saying special relativity has some problems.

    Are you trying to refute special relativity with your theory, or just to understand it better? Have you tried understanding the standard presentations of special relativity, before trying to formulate an alternative version?
     
  14. ralfcis Registered Senior Member

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    This is the Md that shows why the pink light signal has twice the duration of the yellow light signal in this example. I've derived an important formula:

    \(DSR_v= Y_u/Y_w\) where w is the combo of v and u

    as u approaches c, both Y_u and Y_w approach infinity but the infinites cancel out in their ratio which approaches a finite value of DSR_v.

    so if v=3/5c and u = 21523360/21523361c and Y_u =21523361/6561 = 3280 and
    \(Y_w = Y_vY_u(1 + vu/c^2)\) =5/4*21523361/6561(1+12914016/21523361)= 6561

    DSR_v = .5 which is the value Y_u/Y_w approaches when Y_u approaches infinity.

    In this diagram I start off using spaceships instead of light signals to transfer messages between Bob and Alice.

    v is Alice's velocity from Bob at 3/5c
    u is the velocity of Alice's message to Bob at 3/5c (Y_u=5/4), 4/5c (Y_u=5/3), 15/17c (Y_u= 17/8) and c (Y_u= infinity). Hence u approaches c.
    w is the velocity of Bob's message to Alice which is the combo of v and u at 15/17c (Y_w=17/8), 35/37c (Y_w=37/12), 63/65c (Y_w=65/16) and c (Y_w= infinity).

    Notice the progression of Y_u/Y_w to DSR_v = .5 as u approaches c.

    This means the pink light signal is twice as long as the yellow light signal.

    I need to finish this tomorrow.
     
  15. ralfcis Registered Senior Member

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    421
    So you're basically asking me to show you how to plug in numbers from my tables into a formula and you meant to say v=3/5c and u=4/5c and you still have no idea what u_t and v_t are and can't find them in the headers of my tables? Ok if v=3/5, v_t = 4/5 and if u=4/5, u_t=3/5. I didn't even need to look at the tables. Should I take a video of me punching numbers into a calculator? I don't think you're being serious so really if you don't want to put any effort into trying to understand anything I've written, then we should part ways and consider this thread one of the great mysteries in life you have no time for. Like I said, I'm busy.
     
  16. James R Just this guy, you know? Staff Member

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    If
    $\gamma_u=\frac{c}{\sqrt{c^2-u^2}}, \gamma_w=\frac{c}{\sqrt{c^2-w^2}}$
    then
    $\frac{\gamma_u}{\gamma_w} = \sqrt{\frac{c^2-w^2}{c^2-u^2}}$
    In the case where $u\sim c$, that quantity diverges to infinity, unless $w = c$, in which case the quantity is zero, or $w=u$, in which case the quantity has numerical value $1$.

    I don't really understand why you call this DSR_v, unless v is somehow related to w and u.
     
  17. James R Just this guy, you know? Staff Member

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    I now understand that $v_t$ is just related to whatever $v$ is by $c^2=v^2+v_t^2$. I don't understand why anybody would want to use $v_t$ for anything, but that's a separate issue. I can certainly calculate it from the meaningful velocity $v$ if I need it.

    I don't understand why you keep randomly adding and dropping factors of 'c', though.

    Do $v$ and $v_t$ have the same units (metres per second, say), or not?

    If not, then your equation $c^2 = v^2 + v_t^2$ must be incorrect. Do you agree?

    If, on the other hand, $v_t$ does have the same units as $v$ and $c$, after all, then it cannot be the case that
    $v_t=\frac{ct'}{ct}$,
    as you previously stated, because that would be a dimensionless (unitless) quantity.

    No need. Please assume I am able to use a calculator.

    I have asked you some specific questions in my posts above and I have posted many specific comments about what you have written. I have also pointed out several apparent errors in your work. Why are you ignoring the posts where I did those things?

    I have also asked you to clarify some of your definitions so I can better understand you. Will you do that, or not?

    If you don't want to put any effort into explaining your own work, then we should part ways and consider your thread to be one of the great wastes of time in life. I'm busy, too, so if you're not serious we can call this quits now.

    I think you need to decide whether you came here for a discussion, or just to publish your ideas. This is a discussion forum, not your blog.
     
    Last edited: Feb 9, 2021
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  18. ralfcis Registered Senior Member

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    I listed a bunch of forms of this equation including Y=c/sqrt(c2-v2) because v_t =c/Y. So you're saying all the other forms are also dimensionally incorrect?

    I'm not even going to read or answer anything else from you. You keep asking the same questions that I've already answered. No more time for you. However Janus58 is a guy who knows his stuff so I'd be happy to talk to him instead.
     
  19. exchemist Valued Senior Member

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    I've found out why Ralf has come back here: https://www.scienceforums.com/topic/37634-ralfativity/?tab=comments#comment-390894

    He's had his formulae questioned on the nutcase forum and made a Grand Trampling Exit from it. The funny thing is that Write4U, having made a Grand Trampling Exit from this forum, has now shown up there! So we've traded Write4U for Ralf, evidently.
     
  20. James R Just this guy, you know? Staff Member

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    The Y you give here is a dimensionless constant, according to its formula.

    In the other formula you give here, for v_t, it seems that v_t must have the same dimensions as c, because Y is dimensionless. If c is a velocity in metres per second, say, then according to that equation v_t must also be a velocity in metres per second.

    Your "main equation", then, is fine as long as c, v and v_t all have the same dimensions. In that case, I assume you've just made some careless errors every time you have used an example where v_t has no units.

    The only other problem we need to sort out is what your error is in the equation:
    $v_t = ct' / ct = t'/t$, which obviously has v_t as a dimensionless quantity rather than a velocity.

    Is that equation supposed to be correct, or do you want to amend it in light of the above?
     
  21. James R Just this guy, you know? Staff Member

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    Well that's a pity. Here you are, never having found somebody who can understand a Minkowski diagram in 10 years, and you decide to ignore the first person you meet who does understand it.

    I guess you're not really looking for an honest discussion of your "work". Is this more an advertising exercise for you, or something similar?

    Don't be silly. If you answered my questions, obviously there would be no point in my repeating them.

    You've ignored entire posts of mine, as you know. You seem to only be able to deal with tiny snippets at a time.

    That's a pity, because I could probably help you and save you some time.

    Okay.

    Bye, then!
     
  22. ralfcis Registered Senior Member

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    421

    I'll answer this one.
    According to the spacetime diagram v=x/ct which is dimensionless according to you but in the equation v=x/t is used. It's the same with v_t. In the equation it's ct'/t and ct'/ct in the spacetime diagram. In the equation t=Yt' so t'=t/Y and plug that into the equation v_t=ct'/t and you get v_t=c/Y. So v_t can be expressed in distance units and in time units. As I said in an edited post, if you eat 5 hotdogs per hour you're velocity or rate of consumption doesn't involve distance. So if you eat 5 hotdogs per hour and are travelling at 15 miles per hour, how many dogs will be eaten in 2 miles. Oh no, mixed units everywhere!

    I'm glad you gracefully bowed out because I'm supposed to avoid stress after my heart attack and I've seen your arithmetic and algebra capabilities and am sure they match those of your spacetime diagram knowledge and you'll only slow me down and I'm busy. I'll give it a little more time before I leave for Janus to interact but there are too many negatives for me to stay on here. The editor is just too primitive and time consuming to use.

    I've met maybe 3 people who understand relativity and by going from forum to forum I'm hoping to meet more and learn from their observations. This math is constantly evolving.
     
    Last edited: Feb 9, 2021
  23. ralfcis Registered Senior Member

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    421
    Ok I see you put some effort into asking your questions. I just read them and will spend time answering them. I think I'll finish my last post before I do.
     

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