Tony's "quest for knowledge" has spilled over into the forum I call home and I noticed Janus' method for calculating two ships leaving earth at .9c at right angles. If I understand it correctly, he has v = .9c horizontal with a Y of 2.3. He then applies this formula to u perpendicular: u'=u/Y (v's gamma not u's gamma) and comes up with u'=.392. So it's like velocity dilation where v has a perspective of u and the result is u' relative to v. I've never seen this before and am wondering if this method is only applicable to velocities at right angles? Is it derived by breaking up v into t and x and then using time dilation? It also creates a 2 dimensional vector with a resultant value of w= .9817 without an angle mentioned relative to v. I don't know how to use this method if, for example, u was leaving earth at .8c at a 45 degree angle to v=.9c.
Janus' method for non-linear velocity addition
- Thread starter ralfcis
- Start date
Yes I know that formula. I'm interested in how Janus' method works because it may show some deeper short hand understanding. I'm interested if there is such a concept as velocity dilation. I've noticed if I use Yv instead of v in my equations, the equations linearize (use only basic arithmetic) and the infinity inside Y at c, cancels out.
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Janus58
Valued Senior Member
The method I used for the "right angle" scenario works because You don't have to account for length contraction or the relativity of simultaneity.
For example, In the following diagram, we show A and B's velocity as measured from the Earth on top, and then the same scenario as seen from the rest frame of A.
In this case, B always remains directly "above" E at all times. If we were to put a clock (Cb) directly above E and at rest with respect to it, B would move on a straight line between the two. In others words everything in side the dotted box moves to the left as a system from A's perspective. A clock at E and the clock Cb are in sync according to both A and anyone traveling in that box. So if B takes 1 sec to travel 1 sec according to E and Cb (leaves E when both clocks reads 0 and arrives at Cb when both clocks reads 1 sec), According to A, B also leaves E when both clocks read 0 and arrives at Cb when both clocks reads 1 sec. However, according to A, both the clock at E and Cb are time dilated. If the relative velocity between A an E is 0.866c, then according to A, it take 2 sec for the Clock at E and clock Cb to tick off one sec, and thus it takes 2 sec between B leaving E and arriving at Cb. Since the vertical distance between E and Cb is the same for A as it is for E, , this means that the vertical component of B's motion as measured by A would be half that as measured by E. Now it is just a matter of adding the velocity components to get the resultant velocity.
As far as the idea of "velocity" dilation goes. In this case, it is just the consequence of time dilation. Velocity is distance/time. The vertical distance remains the same for both frames, so the measured vertical velocity follows the time dilation.
when the 2 velocities are not at right angles, you have to account for length contraction and relativity of simultaneity, which add extra complications to the calculation.
Attachments
Thanks for taking the time to answer. It'll take me some time to understand it. I think it's percolating in. A sees E's time dilate but B's motion also dilates wrt E so A would see B's dilation through E's dilation. I'll need to calculate to see if I'm getting it.
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Janus, who won the race? Newton or Einstein?
https://photos.google.com/photo/AF1QipPiogT4OFvz8gAvi-LHHCWNFKLEQUk6IGdGcgEw
We want to see the final results of the game and the exact time they arrived. Thanks.
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I will analyze it according to the specific data given by Janus. See if it's a rigorous calculation and a rigorous logical deduction.The scenario I designed is not complicated, so I believe Janus can give the answer directly.
I also give the calculation of the bending of light. The gravitational field of the earth really affects the bending of light, and this influence is decisive and cannot be ignored.
It is hoped that our discussion will affect the future of physics.
So I'll put all my abilities into this discussion. I will be a good match for Janus. Also, for me, he is the best opponent.
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OK, let me go back to my thread.
Viewer;
Janus didn't show enough details, so here is one method, after 4 attempts.
fig.1
A and B are moving at 90 ° relative to E, and relative to each other.
The experiment becomes a closing speed case, with a constant direction between A and B, with E serving only as a reference point.
Relative to the common origin on x, each moves on a 45° course in a 2D plane, without a time axis, i.e. this not a spacetime graphic. The tick marks represent intervals of
x = .9t. Determination of velocity requires 2 separate events, and in this case the common origin serves as the first.
A signal leaves at A1, reflects at B9.4, and returns at A88.4.
The time outbound is calculated using a right triangle by solving for t, with c=1 and v = .90.
t^2 = v^2+v^2(1+t)^2
t = 9.4
fig.2
The return right triangle is proportional to the outbound triangle by a factor or 9.4.
fig.3
After converting distances to times (and defining a um), A calculates relative speed of B as 17.2/17.5 = .98.
Values are only to 1 decimal, sufficient for a fantasy.
Janus didn't show enough details, so here is one method, after 4 attempts.
fig.1
A and B are moving at 90 ° relative to E, and relative to each other.
The experiment becomes a closing speed case, with a constant direction between A and B, with E serving only as a reference point.
Relative to the common origin on x, each moves on a 45° course in a 2D plane, without a time axis, i.e. this not a spacetime graphic. The tick marks represent intervals of
x = .9t. Determination of velocity requires 2 separate events, and in this case the common origin serves as the first.
A signal leaves at A1, reflects at B9.4, and returns at A88.4.
The time outbound is calculated using a right triangle by solving for t, with c=1 and v = .90.
t^2 = v^2+v^2(1+t)^2
t = 9.4
fig.2
The return right triangle is proportional to the outbound triangle by a factor or 9.4.
fig.3
After converting distances to times (and defining a um), A calculates relative speed of B as 17.2/17.5 = .98.
Values are only to 1 decimal, sufficient for a fantasy.
Thank you Phyti for teaching me something very important. I always assumed math was a universal language that can be easily read by mathematicians like in the movies. I'm not claiming to be a mathematician but I assumed since I can read my own math, every mathematician could read it. (I now see no one has ever tried so I have no reason to be so frustrated.) Conversely, I should be able to read your math format. You get the right answer so I should be able to work backwards from that and understand your motivation but I see now it's not so easy. I can't even follow your Minkowski diagrams unless I painstakingly convert them into the format I'm comfortable with. I don't understand most of the lines you draw in and what purpose they serve. I'm not saying your format is wrong, it's just that I can't understand it without putting a lot of effort in. (I totally failed to understand the last one and the accompanying 2 sentences in the length contraction thread.) We just don't speak the same math language.
Now so far as this thread goes, I'm not looking for the right answer, I'm looking for the meaning behind the answer. I try to use the long formula to get the right answer but I have a lot of trouble plugging in numbers and getting the right answer. (For example with two ships leaving earth at a 45 degree angle at .6c, my answer is .513c between them but I'm not confident that's correct.) In the method I've pasted together to try to solve that problem, I think angles in normal life, which use Pythagoras as a sum of squares, are not the same angles as relativity uses which use hyperbolic Pythagoras as a subtraction of squares but I don't know if what I'm saying is true. I have a very simple way of calculating two ships leaving earth at .6c at 0,90 and 180 degrees but my answer at 45 degrees does not match .513c so I'm stuck figuring out why not. I fear that because my math is so different, no one is going to bother to point out the mistake in my reasoning. I'll get the usual answer to stop thinking and just use the wiki formula. I've learned, though, that to get anyone to look at my math, I'll have to describe it in painstaking detail.
Now so far as this thread goes, I'm not looking for the right answer, I'm looking for the meaning behind the answer. I try to use the long formula to get the right answer but I have a lot of trouble plugging in numbers and getting the right answer. (For example with two ships leaving earth at a 45 degree angle at .6c, my answer is .513c between them but I'm not confident that's correct.) In the method I've pasted together to try to solve that problem, I think angles in normal life, which use Pythagoras as a sum of squares, are not the same angles as relativity uses which use hyperbolic Pythagoras as a subtraction of squares but I don't know if what I'm saying is true. I have a very simple way of calculating two ships leaving earth at .6c at 0,90 and 180 degrees but my answer at 45 degrees does not match .513c so I'm stuck figuring out why not. I fear that because my math is so different, no one is going to bother to point out the mistake in my reasoning. I'll get the usual answer to stop thinking and just use the wiki formula. I've learned, though, that to get anyone to look at my math, I'll have to describe it in painstaking detail.
Janus58
Valued Senior Member
Okay, here's how I'd go about visualizing it.
You have A and B starting at the same point and heading off at 0.6c at a 45 degree angle to each other. Imagine a box with one corner at the starting point, and the other where B ends up after 1 sec ( the diagonal of the box will be 0.6 ls long in the box frame, and each side ~0.424 ls) You can also imagine clocks at all four corners.
The top two images show how the start and finish look in the box's rest frame.
The lower two images show the same events as determined by A's rest frame, where A is at rest and the "box" is moving to the left at 0.6 c. According to A both he and B where adjacent to the left bottom clock when it read 0, He also agrees the B arrives at the right upper clock when it reads 1 sec. However, according to A, the box is not a square, being length contracted in the left-right dimension to 0.424 X 0.8 ls = 0.34 ls. Also, all four clocks, due to the Relativity of simultaneity do not all read the same. The clocks on the right side of the box read ~0.25 sec ahead of the left side clock.
This means than in order for B to leave the left lower clock when it reads 0 and arrive at the upper right clock ( the dotted green arrow indicates B's path with relation to the box) when it reads 1 sec, only 0.75 sec tick off on the box clocks according to A. Due to time dilation 0.75/.08 = ~-0.94 sec pass for A in this time.
In that 0.94 sec the box moves ~0.56 ls to the right. B, in moving from left to right relative to the box will displace 0.34 ls to the right, and end up ~0.22 ls to the left of A. It will also be 0.424 ls "above" A.
Thus the distance from A to B along the green arrow is sqrt(-0.22^2+.424^2) = 0.478 ls.
This separation occurred over 0.94 sec as measured by A which gives a relative velocity between A and B of 0.509c A bit off from from what you got with the long equation, but close considering that a lot of the values I used above were rounded out to only 2 significant digits. For example, The time offset due to the relativity of simultaneity is closer to 0.2556 sec rather than 0.25 sec.
I expanded all your numbers and the final answer works out to .5127c. Your last equation of the bottom 1st paragraph should be 0.75/.8=~.94 for anyone else following along. Everything makes sense except I don't understand how the box is a rest frame but inside it, B is going at .6c but A is going at .424c relative to the bottom left corner I assume. I've never seen movement inside a rest frame but I can see if there were, the Newtonian relative velocity formula would apply but not the relativistic combo formula as the velocities are not relative but are closing speeds. Am I interpreting this new trick correctly? If so, this shows any angles between the velocities are handled outside of relativity. I didn't see that coming.
I was approaching this problem by seeing a math pattern for two ships leaving earth at .6c at 0,90 and 180 degrees but couldn't establish any pattern for any other angles. I'll write out the approach in detail. Thanks for your answer..
I was approaching this problem by seeing a math pattern for two ships leaving earth at .6c at 0,90 and 180 degrees but couldn't establish any pattern for any other angles. I'll write out the approach in detail. Thanks for your answer..
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@Janus,
I finally saw your answer to this question. I was recently busy building the Doppler effect model of gravitational field and verifying the program.
Finally, the effort paid off. My model is also correct, and the law of gravity has also been verified on the model.
I will read your answers carefully, and maybe I will continue to give you problems.
I finally saw your answer to this question. I was recently busy building the Doppler effect model of gravitational field and verifying the program.
Finally, the effort paid off. My model is also correct, and the law of gravity has also been verified on the model.
I will read your answers carefully, and maybe I will continue to give you problems.
Janus58, I'm going to butter you up before I get into this. I've only met 2.5 other people like you before, those with a clear (non-wiki) understanding of relativity. I don't think you're a hobbyist because it's difficult to maintain such a clear understanding without being a professional or a teacher in it. So my math is going to be different, it uses uncommon forms of relativistic math. This, as you know, is the primary equation of relativity (I assume the BB code editor is for math script but I'll pass on using it).
(ct')^2 = (ct)^2 - x^2
It has a famous form of
Y= c / sqrt((c-v)(c+v))
but I use my own form
c^2 = v^2 + u^2
where u is the rate of time through time = c/Y = ct/t' which I call the velocity through time.
(For those who can't visualise what a velocity through time looks like, just press fast forward or slow motion on your DVD player. The rate of time you're watching is different from your normal rate of time which is going at the velocity of c through time. c through time is 0 observed velocity through space and c through space is 0 observed velocity through time (which can't be reached))
Therefore c is the sqrt of the sum of squares of 2 velocity components at right angles to each other, v (the velocity through space) and u (the velocity through time). Everything moves at this composite velocity of c but, just like any velocity addition in relativity, the faster you are observed to move through space, the slower you are observed to move through time because no combination of any type of velocities can exceed c.
There are 3 main ways to graphically depict relative velocity as rotations of cartesian coordinates: Minkowski, Epstein and Loedel diagrams. The Epstein is a true rotation that represents a circular sum of squares relationship between the 2 coordinate systems whereas Minkowski is a hyperbolic difference of squares representation. Imagine two sheets of graph paper; one with ct and x axes and the other with ct' and x' axes and you rotate one sheet over the other pivoting at the common origin. The Epstein diagram rotates the ct/x sheet clockwise over the ct'/x' sheet whereas the Minkowski diagram rotates the ct'/x' sheet clockwise over the ct/x sheet. (Seems like this small difference should not create such vastly different graphical representations.) Minkowski also folds its x' axis over the x axis so all representations of c overlap the same 45 degree line which is graphically convenient to make c look the same from all perspectives. The Epstein diagram doesn't use this trick so things get messy with the slope of c. Just a note to all the Wiki readers, the Minkowski rotation is described as a clockwise rotation of ct' and a counter-clockwise rotation of x' so my mathematical interpretation is in quite a different form but still valid. I go further in my own Minkowski-hybrid diagram in that I need no rotation or mirror flip of the x-axis (meaning length is mathematically invariant) to get the correct results for any relativistic problem. (It's just a different form of math and it works so it's valid.) I also use Loedel perspectives and Loedel simultaneity (non-relativistic terminology) on my Minkowski-hybrid extensively to expose the underlying objective reality of proper time and proper space as opposed to the subjective "reality" of perspectives. But more on future threads if there are any.
All this will become clearer when I show my approach to solving the two ships leaving earth at different angles problem. Hopefully when I get to the bottom I'll be able to solve the 45 degree problem only using relativistic concepts without Pythagoras.
(ct')^2 = (ct)^2 - x^2
It has a famous form of
Y= c / sqrt((c-v)(c+v))
but I use my own form
c^2 = v^2 + u^2
where u is the rate of time through time = c/Y = ct/t' which I call the velocity through time.
(For those who can't visualise what a velocity through time looks like, just press fast forward or slow motion on your DVD player. The rate of time you're watching is different from your normal rate of time which is going at the velocity of c through time. c through time is 0 observed velocity through space and c through space is 0 observed velocity through time (which can't be reached))
Therefore c is the sqrt of the sum of squares of 2 velocity components at right angles to each other, v (the velocity through space) and u (the velocity through time). Everything moves at this composite velocity of c but, just like any velocity addition in relativity, the faster you are observed to move through space, the slower you are observed to move through time because no combination of any type of velocities can exceed c.
There are 3 main ways to graphically depict relative velocity as rotations of cartesian coordinates: Minkowski, Epstein and Loedel diagrams. The Epstein is a true rotation that represents a circular sum of squares relationship between the 2 coordinate systems whereas Minkowski is a hyperbolic difference of squares representation. Imagine two sheets of graph paper; one with ct and x axes and the other with ct' and x' axes and you rotate one sheet over the other pivoting at the common origin. The Epstein diagram rotates the ct/x sheet clockwise over the ct'/x' sheet whereas the Minkowski diagram rotates the ct'/x' sheet clockwise over the ct/x sheet. (Seems like this small difference should not create such vastly different graphical representations.) Minkowski also folds its x' axis over the x axis so all representations of c overlap the same 45 degree line which is graphically convenient to make c look the same from all perspectives. The Epstein diagram doesn't use this trick so things get messy with the slope of c. Just a note to all the Wiki readers, the Minkowski rotation is described as a clockwise rotation of ct' and a counter-clockwise rotation of x' so my mathematical interpretation is in quite a different form but still valid. I go further in my own Minkowski-hybrid diagram in that I need no rotation or mirror flip of the x-axis (meaning length is mathematically invariant) to get the correct results for any relativistic problem. (It's just a different form of math and it works so it's valid.) I also use Loedel perspectives and Loedel simultaneity (non-relativistic terminology) on my Minkowski-hybrid extensively to expose the underlying objective reality of proper time and proper space as opposed to the subjective "reality" of perspectives. But more on future threads if there are any.
All this will become clearer when I show my approach to solving the two ships leaving earth at different angles problem. Hopefully when I get to the bottom I'll be able to solve the 45 degree problem only using relativistic concepts without Pythagoras.
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Janus58
Valued Senior Member
I've never seen movement inside a rest frame but I can see if there were, the Newtonian relative velocity formula would apply but not the relativistic combo formula as the velocities are not relative but are closing speeds. Am I interpreting this new trick correctly? If so, this shows any angles between the velocities are handled outside of relativity. I didn't see that coming.
[/quote] A "rest frame", is just short hand for " the reference frame where the object at interest is measured as being at rest". This does not mean other objects in motion relative to this object are not also "in" that same frame, they are just not at rest when measured in that frame. For this scenario, there are an infinite number of inertial frames of reference that these objects are "in". It is just that certain ones are easier to work things out from than others.
I was approaching this problem by seeing a math pattern for two ships leaving earth at .6c at 0,90 and 180 degrees but couldn't establish any pattern for any other angles. I'll write out the approach in detail. Thanks for your answer..[/QUOTE]
Ralf;
Your reply was surprisingly pleasant. (Wondering, maybe he hired a publicity agent.)
Long long ago, in school, I noticed math wasn't the most popular subject, and those who studied it, did so with varying degrees of difficulty. It wasn't that they couldn't understand it. The school system expected everyone to graduate yearly in a one size fits all plan.
I think you are one of those people who enter the theater in the middle of the movie. Unaware of what has transpired, you can't connect the current events with those that have already occurred. You seemed to have learned bits and pieces from random sources without an integrated program of learning.
Or, you could be one of the self taught musicians, who know the essentials of music and your instrument, but can't read music. You can only play songs you know, whereas those with formal training can play songs they don't know.
Math is standardized, so you have to know its definitions. A little history.
Prior to the 1600's, light propagation was thought to be instantaneous, which implies a universal time. If you saw a star grow brighter, it was happening while you watched.
At that time, if a scientist had a laser-like device and sent a beam to the moon, it would look like the 1st graph. Signal sent at 1 sec. and return at 3.5 sec. He would obviously wonder, why the 2.5 sec. delay. One explanation might be, the light interacts with matter before it returns. As more experiments were done with objects at greater distances, someone would detect a pattern where the delay increased with increasing distance. Delay in light transit time is what actually occurred in astronomical observations. Refer Ole Romer, 1670. Even if this new discovery was included in the graph, it would still look the same.
The problem is scale. The unit of time is 1 second, with the unit of space 3(10)^8 meters. Enter Minkowski, and as a mathematician, he seeks to generalize. If t is multiplied by c, then ct can be treated (mathematically) as a 4th dimension, just another line on paper. This scaling factor simplifies SR in calculations, and enables sensible graphical comparisons in the near light speed range. The horizontal x axis is now object motion vt, and the vertical ct axis is light motion, as in the 2nd graph. The ct axis U, is standard as the reference frame. The 'time/world line' of an object has a slope of vt/ct = v/c, a speed profile. Thus light has a slope of 1 or 45 deg.
In the example, B has a speed of .6 relative to U. The red calibration curve (Max Born) is typically used to indicate time dilation, .8 in this case.
If you know algebra, then 1/gamma = sqrt(1-a^2), with a = v/c. 1/gamma also equals the circular arc from ct = 10 to ct = 8. If you have a drawing app, you can copy and paste the red curve in as needed, but it's much simpler to draw an arc. Then the gray line between the ends of the red curve can substituted for the curve to transfer any B-time to the U-time scale, eliminating all those tick marks. As you see, there are practical shortcuts available, especially when there are only two frames involved.
My personal preferences are blue for light, light gray for measurements and construction lines, and other colors as needed. Strive for simplicity. Any measurement requires a round trip light signal (blue). I don't use a grid since the CAD system makes measurements to great precision, and only a few are necessary.
In the example,
U assigns coordinates (6, 10) to B.
U can transform his coordinates via the LT, to those of B as (0,8).
The spacetime graphic is a geometric application of the LT.
B perceives he is at (4.8, 8) relative to U.
His perception is what he thinks/concludes after a complete analysis of images, calculations, etc.
Your reply was surprisingly pleasant. (Wondering, maybe he hired a publicity agent.)
Long long ago, in school, I noticed math wasn't the most popular subject, and those who studied it, did so with varying degrees of difficulty. It wasn't that they couldn't understand it. The school system expected everyone to graduate yearly in a one size fits all plan.
I think you are one of those people who enter the theater in the middle of the movie. Unaware of what has transpired, you can't connect the current events with those that have already occurred. You seemed to have learned bits and pieces from random sources without an integrated program of learning.
Or, you could be one of the self taught musicians, who know the essentials of music and your instrument, but can't read music. You can only play songs you know, whereas those with formal training can play songs they don't know.
Math is standardized, so you have to know its definitions. A little history.
Prior to the 1600's, light propagation was thought to be instantaneous, which implies a universal time. If you saw a star grow brighter, it was happening while you watched.
At that time, if a scientist had a laser-like device and sent a beam to the moon, it would look like the 1st graph. Signal sent at 1 sec. and return at 3.5 sec. He would obviously wonder, why the 2.5 sec. delay. One explanation might be, the light interacts with matter before it returns. As more experiments were done with objects at greater distances, someone would detect a pattern where the delay increased with increasing distance. Delay in light transit time is what actually occurred in astronomical observations. Refer Ole Romer, 1670. Even if this new discovery was included in the graph, it would still look the same.
The problem is scale. The unit of time is 1 second, with the unit of space 3(10)^8 meters. Enter Minkowski, and as a mathematician, he seeks to generalize. If t is multiplied by c, then ct can be treated (mathematically) as a 4th dimension, just another line on paper. This scaling factor simplifies SR in calculations, and enables sensible graphical comparisons in the near light speed range. The horizontal x axis is now object motion vt, and the vertical ct axis is light motion, as in the 2nd graph. The ct axis U, is standard as the reference frame. The 'time/world line' of an object has a slope of vt/ct = v/c, a speed profile. Thus light has a slope of 1 or 45 deg.
In the example, B has a speed of .6 relative to U. The red calibration curve (Max Born) is typically used to indicate time dilation, .8 in this case.
If you know algebra, then 1/gamma = sqrt(1-a^2), with a = v/c. 1/gamma also equals the circular arc from ct = 10 to ct = 8. If you have a drawing app, you can copy and paste the red curve in as needed, but it's much simpler to draw an arc. Then the gray line between the ends of the red curve can substituted for the curve to transfer any B-time to the U-time scale, eliminating all those tick marks. As you see, there are practical shortcuts available, especially when there are only two frames involved.
My personal preferences are blue for light, light gray for measurements and construction lines, and other colors as needed. Strive for simplicity. Any measurement requires a round trip light signal (blue). I don't use a grid since the CAD system makes measurements to great precision, and only a few are necessary.
In the example,
U assigns coordinates (6, 10) to B.
U can transform his coordinates via the LT, to those of B as (0,8).
The spacetime graphic is a geometric application of the LT.
B perceives he is at (4.8, 8) relative to U.
His perception is what he thinks/concludes after a complete analysis of images, calculations, etc.
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