Relativity and simple algebra II

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

  1. Michael 345 New year. PRESENT is 71 years old Valued Senior Member

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    Ummmmm So I posted (explained) to you the details you replied with were confusing to me and you come back with some post which makes even less sense

    I have to go with troll

    James seems to be holding on (out) that you might have something (????? what)

    Will leave you two alone

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  3. ralfcis Registered Senior Member

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    Take two atomic clocks, move them slowly apart, measure light speed between them while they're still moving (to not disrupt clock sync by maintaining constant relative velocity) and call me in the morning. Clearer now?
     
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  5. Michael 345 New year. PRESENT is 71 years old Valued Senior Member

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    Sure

    This is more so



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  7. ralfcis Registered Senior Member

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    Can you not see the difference between how I slowly separated the clocks at a constant velocity that was maintained during the measurement and how he moved one clock and stopped it? He invoked desync and permanent age difference between the clocks through the twin paradox and I maintained sync by maintaining constant relative velocity. Now go look that up somewhere where someone says that's not a valid measurement of the one way speed of light and I'll show you someone who doesn't understand relativity.
     
  8. arfa brane call me arf Valued Senior Member

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    That's what you think. You can't actually show this, according to Einstein.
     
  9. ralfcis Registered Senior Member

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    Time dilation is reciprocal during constant relative velocity. Permanent age difference or desync between the clocks happens during the twin paradox which I avoid by not making any velocity changes in my test set up. You can show this in a Minkowski diagram or even better in a Loedel diagram as I've shown on this thread. You can calculate it or just use my Loedel perspective which cancels out the hysteresis of perspective simultaneity and is a direct window into proper simultaneity. The experimental results will not differ from the expected results. Even better using the Loedel set up because the light leaves from the center so there's no chance that direction of the light can cause 0c in one direction and c in the other. You have to be aware of the basics of relativity, not what you tube or wiki tells you.
     
  10. ralfcis Registered Senior Member

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    The more I think of it, the more I think this is a perfect example of the power of Loedel proper time simultaneity over relativity's perspective time simultaneity.

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    Here's the Loedel diagram of two atomic locks leaving Earth at 1/3c in opposite directions. In a lab set up you would want the two clocks separating as slowly as possible from the third stationary clock between them which represents the stationary Earth frame Loedel half speed velocity =0. The Loedel stationary frame allows a glimpse of proper time simultaneity of the two separating clocks. This means the green lines of Loedel simultaneity share the same proper times of the moving clocks whereas the perspective lines of simultaneity would connect two different perspective times on both ends.

    So even at 1/3c separation from the Earth frame it is possible to measure the one way speed of light in both directions. Each ship would have star charts of its distance from earth and their on-board clocks would be programmed to send out light signals at t=2 proper time within their own frames. They would receive the light signals at t=4 proper time within their own frames. Since it's proper time they don't need the other ship's perspective time. In a lab, of course, the two clocks would register that they indeed sent out and received the signals proper time simultaneously (they were all sync'd to t=0 at the start) and they knew their proper distance separation at those transmit and receive times. The reference frame itself does not share the same proper time as the other two clocks but its time is calculable using relativity. It's not even relevant at slow speeds because it would not be measurably different from the proper times on the other two clocks.

    In a lab set up the slopes of the blue lines would be near vertical and would need several hours of slow clock movement to achieve a decent separation. Also, the stationary clock would need to be programmed (its proper time would not be measurably different from the other clocks' proper times) to send out the light signal to both. The distance would be known and the time is proper time received minus proper time transmitted. The proper time received would be exactly the same on both moving clocks (which remain moving throughout) and if you're a real stickler for accuracy, the proper time on the stationary frame could be adjusted for the slow velocity of the moving clocks. This experiment is designed to verify relativistic effects, not to discover them.
     
    Last edited: Feb 18, 2021
  11. James R Just this guy, you know? Staff Member

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    ralfcis:

    Thanks for your replies to post #71 and #72. I'll comment on those next. But this post is a reply to post #127.

    Following your replies to #71/72, I think I now understand what you mean when you refer to Loedel diagrams and Loedel lines.

    To me, your Loedel diagrams are essentially the same as Minkowski diagrams. One addition you make is to label the proper times of the "moving" frames on the worldlines of the objects/observers who are moving. The other main addition you make is to add "Loedel lines", like the green ones on your graph in #127, which appear to be just straight lines that connect equal "proper times" of selected "moving" observers.

    I see that you have noted that, on your diagram here, the proper time of the Earth or lab frame is different from the proper time of either of the moving clocks. Your green "Loedel line" at proper time t=2 does not pass through the t=2 position on the coordinate axis (i.e. for the Earth frame). In other words, that "Loedel line" does not connect the proper times of all three observers (the two moving clocks and the stationary Earth), but only two out of the three.

    You have also drawn the hyperbola that corresponds to the equation
    $s^2=(ct')^2=(ct)-x^2$.
    In relativity, this would be a curve of constant $s$ (call it 'the spacetime interval', maybe).

    In fact, for your two moving clocks there, there are two such hyperbolas, but due to the symmetry of this scenario, with the two moving clocks have velocities $\pm c/3$ in the Earth frame, they happen to coincide on this diagram.

    Those hyperbolas are, I think you would say (?), curves of "equal proper time".

    I don't really understand what use your straight green "Loedel lines" are, since only the endpoints of those lines have "equal proper time", as I'm sure you'll agree. Intermediate points have no special significance, as far as I can see. Is that correct, or do intermediate points on those straight green lines mean something in particular to you?

    The light signals you have drawn are each emitted at proper time t=2 on the "moving clocks", and are received at proper time t=4. As we see from the diagram, the lines are symmetrical, which is again due to the $\pm c/3$ velocities of the two moving clocks (i.e. they both have the same speed, just in opposite directions relative to Earth).

    What does not appear on your diagram are any lines of simultaneity that would show us which events occur simultaneously in either of the moving clock frames. Note: I am not talking about "proper time simultaneity" - those would be the curves of "equal proper time" discussed above - but the times which all the (synchronised) clocks in the relevant reference frame would agree were simultaneous (i.e. clocks everywhere display the same time reading, in that frame). Obviously, any horizontal line drawn on the graph is a line of simultaneity for the Earth frame, but lines of simultaneity in either of the two "moving clock" frames would not be horizontal. They would also not be parallel for the two moving clocks because those two clocks are moving in opposite directions, away from one another.

    We could ask a question like this, for example: for an observer in the frame moving at $v=+c/3$, at what time did the clock moving at $v=-c/3$ emit its light signal? The answer to that question would not be "At $t'=2$", where t' is the time coordinate in the frame of the $v=+c/3$ clock; it would be a time coordinate with $t'>2$ in that frame. In the same frame, the answer to the question "At what time was the signal received?", the answer would be $t'=4$, of course, because that event occurs at $x'=0$, whereas the sending of the signal happened at some $x'<0$ coordinate -i.e. at a distance from the $v=+c/3$ moving clock.

    The elapsed coordinate time in the frame of the $v=+c/3$ moving clock, between the emission of the signal from the $v=-c/3$ clock and the receipt of that signal, is less than 2 seconds. Note: this would be what an observer in the frame of the $v=+c/3$ clock would actually measure. But it is also the case that that observer would measure the distance the light had to travel from the other clock to be less than 2 light-seconds (or light-years, or whatever distance units we're using). Thus, when that observer calculates the speed of the light, the calculated value would be $c$, as usual.

    With all of this in mind, there's not a lot more to say about the this "Loedel diagram". It might be worth mentioning that, from the $v=+c/3$ clock's perspective, the light signal from the other clock was emitted after $t'=2$, while in the $v=-c/3$ clock's perspective, the light signal was emitted at $t''=2$, where the double-prime indicates the $v=-c/3$ frame's clocks, which of course are not synchronised with the $v=+c/3$ frame's clocks.

    To summarise all of this: there's nothing "wrong" with the spacetime diagram (call it a "Loedel diagram" if you like; it's just a label). It doesn't tell us everything we might like to know about the situation, especially from the (different) perspectives of the two moving clocks, but that's okay.

    The straight green "Loedel" lines only connect the "proper times" at the two ends. SR lines of simultaneity (I don't know why you want to use the term "perspective lines") have the property that every point on the line is an event that happens simultaneously in whichever frame the line of simultaneity applies to.

    Not a problem. You'd have to bear in mind that the star charts would have been drawn up using Earth rulers, of course. What you say about proper times here is 100% correct.

    Correct, if all they are interested in is when they sent their own signal and when they received the other one, since both of those events obviously happen at this ship, not at the other one.

    More accurately, an observer in the lab frame would agree that the two "moving" clocks sent out their signals at the same (Earth) time, and, at that time, both displayed the same "proper time" readings, according to the Earth observer.

    Note that all three clocks have different "proper times". We can see that all three time axes head in different directions on the diagram. It is incorrect to say that the two "moving" clocks share the same "proper time". They do not. What is true is that they both appear to tick at the same rate according to the Earth observer and always display the same times according to the Earth observer.

    If we consider, for example, the clock moving at $v=+c/3$, any observer in the rest frame of that clock would see the $v=-c/3$ clock as ticking slower than the $v=+c/3$ clock, which is the reason why, in the $v=+c/3$ observer's frame, the $v=-c/3$ clock emits its light signal later than $t'=2$, rather than at $t'=2$.
     
    Last edited: Feb 19, 2021
  12. ralfcis Registered Senior Member

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    no nothing in particular
     
  13. ralfcis Registered Senior Member

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    no importance except for relativity. I use proper time from which perspective times are derived. I can do all these diagrams using primarily perspective times and I can see how relativty's math works with them but they have no importance to me. At no time do I care what the others perspective time is because I have my own clock and invariant length which gives me all the info I require independent of the other guy.
     
  14. James R Just this guy, you know? Staff Member

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    You decide what's important to you, of course.

    In real life, people are often interested in the time ordering of events (i.e. which event happened first), or in whether two events were simultaneous (happened at the same time).
     
  15. ralfcis Registered Senior Member

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    Simultaneously from a perspective but I use proper time simultaneity from the Loedel proper time simultaneity which does not exist in relativity. When Bob's t=5 joins Alice's t'=4, those are not the same times on his line of simultaneity. In the loedel lines of simultaneity, the only time the endpoints are not the same proper time is after a velocity change in the twin paradox and the two proper times at the endpoints are not the same value permanently until the next velocity change. The slopes of the green lines change during the time of relative velocity imbalance until a new balance is achieved. This is how Alice ends up aging less than Bob permanently.
     
  16. ralfcis Registered Senior Member

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    They do because within their own frames time beats at the normal proper time rate of c. The loedel lines of simultaneity allows the loedel perspective to see this. Bob and Alice can't because they can only see from their own perspectives. However, they can use light signals to and from the Loedel perspective to see this proper time simultaneity after the fact.
     
  17. ralfcis Registered Senior Member

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    There is no way to determine time ordering in relativity because of relativity of simultaneity from different perspectives. But being able to peer into the simultaneity of proper times which only exist on the on-board clocks allows one to state what was the first event based on proper time. Like I said, the sun could disappear and we can't know that for 8 minutes but there's no perspective that can come before the proper time perspective which is the source of causality and the Earth's inescapable but eventual destruction.
     
  18. James R Just this guy, you know? Staff Member

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    ralfcis:

    I'm now going to quickly reply to your responses regarding posts #71/72.

    Your responses clarified a lot of things for me about your approach.

    You're right that some questions in there were "AA" (already answered). Unless there's disagreement, or I have follow-up questions, I won't say anything more about those.

    Is that what you're doing? Proposing alternative math to special relativity?

    To me, it looks like you're mostly using the equations of special relativity, as they apply to certain "special case" situations.

    Also, it seems to me that your calculations of the speeds of light signals don't always preserve the constancy of the speed of light. When they don't, you seem to want to explain it away as having something to do with Doppler shifts or other things (like "proper velocity", for instance).

    In a lot of cases, your calculations are correct, and I applaud you for that. Over the years, we have had a lot of people come to this forum claiming to have "disproved" relativity, who have been utterly unable to do the maths properly, let alone to understand what the maths is telling them. It seems to me that you have taken the time to study the subject and to gain some understanding of it. If you've done that all on your own, without being formally taught by anybody, that's great.

    Well, no. If I stand on a football field and kick a football, obviously the velocity of the football in the Earth frame is non-zero while it is in the air. How, if I run along the football field and watch the football in flight, then in my running frame (which is not the same as the Earth frame), the football has a different velocity to what it has in the Earth frame, but in both frames its velocity is non-zero. This is exactly the kind of situation the velocity-addition formula is made for:
    $u'=\frac{u-v}{1-\frac{uv}{c^2}}$.
    Here, $u$ might be the velocity of the football in the Earth frame, $u'$ is its velocity in my running frame, and $v$ is the speed of my running frame, relative to the Earth frame.

    I get it now. Your Loedel lines are just visual indicators that show where certain points of "equal proper time" are. Really, only the endpoints of those lines are meaningful (thanks for clarifying that), and no other points on those lines are important.

    Got it. Thanks.

    Technically, "proper time" is the time, in a particular frame, between two spacetime events that happen at the same spatial coordinate in that frame.

    What do you mean by "co-located frames"? Frames that have no velocity relative to one another? And what "value" are you referring to, that might not be the same?

    Relativity has no problem dealing with proper times; it doesn't "forbid" any consideration of them.

    My impression (correct me if I'm wrong) is that you think that "equal" proper times in different frames are somehow "connected", or comparable to one another, which might be why you want to connect them with Loedel lines. From my (SR) point of view, points of equal $s$ are "connected" by the invariant relationship (in one dimension)
    $s^2=(ct')^2-x'^2 = (ct)^2-x^2$.
    You use a special case of that and claim to be connecting "proper times" for events that have $x'=0$. That's not exactly what's going on there.

    Yv seems to cause you all sorts of problems, because when you use it, you end up calculating light speeds that aren't equal to $c$. I've seen you claim that some observer could actually "measure" the speed of a light signal to be other than $c$, based on a faulty analysis using Yv.

    There's no notion of "proper simultaneity" in SR because proper times intervals in different frames are not directly comparable with one another. Clocks run at different rates in any two frames that are moving relative to one another, from the perspective of either one of those two frames.

    But then she's using rulers from Bob's frame, rather than her own rulers.

    Yes she can, in principle. Her destination would physically appear closer to her, if she was moving fast enough to notice (3/5c would do the trick).

    No. If she put a whole bunch of her metre rulers end to end, the number of metres she'd measure between Earth and the destination planet or "finish line" would be less than what Bob would measure with his rulers, in your example.

    You should talk in terms of spacetime events, which are "points" with space and time coordinates. A "length" or "distance" is the difference between the spatial coordinates of two specified events, and a "time interval" or "duration" is the difference between the time coordinates of two specified events. And, of course, we can also talked about the "spacetime interval", which is the value of $\Delta s$ calculated according to the usual formula, for the two events. This is the standard terminology. It's very specific and it avoids confusion and misunderstanding.

    You're not saying the Lorentz transformations are incorrect, though. Or are you?

    If you are claiming they are incorrect, then you'll need to go back and specify which of the two postulates of special relativity you want to modify or throw away, because the Lorentz transformations follow from those postulates.

    By the way, I notice that you didn't comment at all on my analysis in post #72, where I used the Lorentz transformations to examine the light signal. Do you agree with my analysis there, or not? If not, which part(s) are incorrect?
     
    Last edited: Feb 19, 2021
  19. James R Just this guy, you know? Staff Member

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    (continued...)

    I'm glad you clarified that. So you accept that length contraction is required in SR.

    It sounds like you don't believe that length contraction is a real effect, even though you accept time dilation. Is that a correct summary of your views?

    Do you think that you can make or that you have made a self-consistent theory of relativity that assumes that lengths are invariant in all frames?

    If she's using Bob's star charts, then she's using Bob's rulers. Agreed?

    Do you agree with me that, if she laid out one of her own metre rulers next to Bob's (as she watches Bob's one fly past), she'd measure two different lengths for the two rules, even though the rulers would be identical if brought to rest with respect to one another?

    Or do you disagree?

    I showed your explicitly how the relevant events transform between the two frames in post #72, using the Lorentz transformations.

    Maybe it would be useful if you could specify the spacetime coordinates of those same two events in the two frames, according to your calculations/theory. Then we can compare.

    If her velocity was "wavering" rather than constant, then we'd be drawing a much more complicated spacetime diagram, wouldn't we? Using your method, you'd be having a much harder time marking off the "proper times" on the wavering worldline.

    In principle, if we know her spacetime coordinates in any single frame (say Bob's frame), then we can easily calculate them in any other given frame, so it's not "insane". It's just a matter of doing some math. You can't pretend your job would be any easier than mine.

    You avoided my question, there.

    We agree that clocks in different frames do not remain synchronised with one another, right? If they did, there could be no displayed time differences at the end of a one-way or two-way trip (like in the "twin paradox".

    You seem to want to say that their "proper times" remain synchronised in some way, and you try to emphasise that by drawing your "Loedel lines". But, as I have pointed out many times, those lines aren't lines of simultaneity.

    That's not very fair.

    I haven't "redacted" anything significant, in responding to your posts. On the contrary, I have been putting a lot of effort into trying to understand your arguments. It's not because I'm stupid, or because I don't want to hear anything that isn't "standard SR". Mostly, it's because your explanations haven't been very clear, and your use of terminology isn't always "standard".

    Notice that I have done nothing but "plain old algebra", just like you. It's not like doing the math using the Lorentz transformations is difficult.

    You don't seem to be brave enough to set out, in a simple way, how your "theory" of relativity differs from SR, or even if it differs. I've asked you several times whether you believe you have just discovered a mathematically simpler way of analysing situations in relativity, or whether you believe you've found errors or deficiencies in SR. You keep avoiding that question. It's almost like you're not sure, yourself.

    Try not to get angry. You're sounding upset.

    I agree with you that there are relativity-of-simultaneity effects. I've discussed some of them in detail.

    You can't honestly claim that I'm "just not getting any of this at all". Have I made a mistake about something? I've tried to be very careful in asking you questions when I'm not sure what you're saying. You seem to get angry when I ask you questions, like it's a waste of your time to try to answer me, or something.

    We don't have to have this discussion, if you think you have better ways to use your time. Just let me know if you want to stop.

    Remember, that was back at post #71. There have been 50 more posts since then. I now understand that you were referring to lines of equal $s$ (spacetime invariant parameter), in effect.

    We might have saved a few of those 50 extra posts if you'd responded to posts #71/72 at the time I asked those questions.

    I don't like pulling rank by citing my academic qualifications in discussions like these. My arguments should stand and fall on their merits, not by the number of university degrees I have, or whatever.

    Please assume, for future reference, that I have more than the "basic popular understanding" of relativity, if that helps.

    It is clear to me that you have some understanding of relativity, which is good, because like I said, we get a lot of self-styled backyard scientists here who don't have a clue what they're talking about. I'm trying to get a handle on what you might not understand, as well as on what you do understand.

    In this case, the thing we're measuring is a bunch of spacetime events. Those events themselves are not variable. Everybody, in whichever frame you like, agrees on what happens - i.e. which events occur in spacetime. But observers in different frames certainly do not agree on the space and time coordinates of any given event. Relativity is all about translating their different coordinates from one frame to another.

    You and I seem to agree that two observers who use identical clocks can have those two clocks tick off different amounts of time between a specified pair of spacetime events. The events weren't variable; both observers agree on them. But the measured time between the events is certainly variable, relative. It depends on who is doing the measuring, even with identical equipment.

    Similarly, the measured lengths of objects are variable, even when the observers are using identical rulers.
     
  20. James R Just this guy, you know? Staff Member

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    A simple way to remember the rules is:
    • moving lengths contract
    • moving clocks run slow
    That is, anybody who tries to measure the length of something that is moving relative to them will measure a shorter length (in the direction of motion) than they would if it was stationary relative to them. The length could be the length of an object like a spaceship or a banana, or is could be something like the distance between two planets that are moving relative to the observer.

    Similarly, anybody who observes a clock that is moving relative to them will calculate/see that the moving clock is running slower than an identical clock that is stationary relative to them.

    I don't understand what you're saying about "the reference frame's perspective through the moving frame's perspective". A reference frame essentially is a "perspective": formally, its a collection of stationary clocks and rulers held by some observer and used to give spacetime coordinates to events. I defined this in more detail in a previous post.
     
  21. James R Just this guy, you know? Staff Member

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    That's an unhelpful distraction from the discussion I'm trying to have with ralfcis, if you ask me.

    If you're really interested in having that discussion with him, and he with you, then go for it. I'm not going to get involved. Not in this thread, anyway. It has nothing to do with the cause of any concerns ralfcis is having about relativity.
     
  22. James R Just this guy, you know? Staff Member

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    You're right that two observers in different frames won't always agree on which of two events happened first, or whether they both happened simultaneously. However, the matter of causality is handled nicely in relativity. We've already talked about the spacetime interval:

    $\Delta s^2 =(c\Delta t')^2 - (\Delta x')^2=(c\Delta t)^2 - (\Delta x)^2$

    For two given events, if $s^2 >0$ then it is possible for one event to cause the other. If $s^2 < 0$ then it is impossible for either of the events to cause the other. And if $s^2=0$, then a light signal can be sent from one event to the other, carrying information that can cause change. Since the quantity $s^2$ is a spacetime invariant (has the same value in every reference frame, for the same two events), everybody in all of spacetime, no matter what their state of motion, agrees on whether two given events can be causally connected.

    So, if Event 1 is the Sun exploding, and Event 2 is the Earth being burnt to a crisp, then $\Delta s >0$ for those two events, which means that one of them can cause the other. In the Earth frame, Event 1 is observed occur before Event 2 and to cause it, so Event 1 must cause Event 2 in every frame of reference, and no frame can see Event 2 happen before Event 1.

    On the other hand, if Event 1 is "I'm eating a sandwich in Australia at 8 am UT" and Event 2 is "You're making a sandwich in Florida at 8 am UT", then $\Delta s <0$ for those two events, which means that it is impossible for either event to cause the other. There are reference frames in which observers will say I ate my sandwich before you made a sandwich, and other frames in which observers will say you made a sandwich before I ate my sandwich, but all will agree that neither one of those events could affect (or cause) the other.

    Note that it is possible for very similar events to have different causal relationships (kind of obvious). If Event 1 is "I'm eating a sandwich in Australia" and Event 2 is "You're mailing a sandwich to Australia from Florida", then obviously those two events could possibly be causally connected, depending on the timing and the precise distance between the two events.
     
  23. ralfcis Registered Senior Member

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    I wrote a response but let's just continue with the way things are going. I will continue to be guarded.
     

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