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Time dilation - No Need to Do the Math

Time dilation - No Need to Do the Math - Physics Forum

Time dilation - No Need to Do the Math - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #11  
Old 10-30-2003, 05:31 AM
Stephen Bint
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Default Time dilation - No Need to Do the Math


"Bonnie Granat" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
It doesn't matter whether you are staiuonary, or moving, with respect to
anyone or anything else. I am moving at 10kph with respect to you. It could
be that you are travelling along a road at 45kph and I am overtaking you at
55kph. Or you could be standing at the side of the road while I pass you in
a car doing 10kph.

The "v" in the time dilation equation refers only to the observed object's
speed with respect to the observer. Our speeds with respect to everything
else have no effect on the outcome of the calculation of time dilation.



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  #12  
Old 10-30-2003, 09:43 AM
Dirk Van de moortel
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Default Time dilation - No Need to Do the Math


"Stephen Bint" <[Only registered users see links. ]> wrote in message news:3fa07dad$0$118$[Only registered users see links. ].. .

Troll alert:
[Only registered users see links. ]

Dirk Vdm


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  #13  
Old 10-30-2003, 09:45 AM
Martin Hogbin
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Default Time dilation - No Need to Do the Math


"Stephen Bint" <[Only registered users see links. ]> wrote in message news:3fa0893c$0$122$[Only registered users see links. ].. .
..>

There is a difference. Dan's view is supported by thousands
of physicists throughout the world who have studied the subject
in great detail for nearly a century.

You should start by understanding how two inertial observers
in relative motion would measure one another's clocks. It
would then be useful to understand how they would _see_
one another's clocks, which is different.

You have a choice. You can either join the list of crackpots
who frequent this group, and be treated with scorn and derision
by the physicists here, or you can make an effort to understand
the subject.

Martin Hogbin



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  #14  
Old 10-30-2003, 09:50 AM
Dirk Van de moortel
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Default Time dilation - No Need to Do the Math


"Martin Hogbin" <[Only registered users see links. ]> wrote in message news:bnqmjp$3q7$[Only registered users see links. ]...

He has been here before with another name.
Same troll, new name.

Dirk Vdm


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  #15  
Old 10-30-2003, 01:55 PM
sal
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Default Time dilation - No Need to Do the Math

I disagree with the subject. If you don't do the math you can never
understand the heart of the problem, which is clock synchronization.
(Well, that's _one_ heart of the problem, anyway...)

I posted some of the math for the twin paradox in your earlier thread,
but I only posted it on alt.sci.physics. So, I've taken this
opportunity to spruce it up by adding the affine transformations to
rezero the clocks.

Others have addressed the issue of what each twin sees through a
telescope, so I haven't done that here. I also didn't take the extra
step of filling space with "virtual observers" and reporting what they
see. That's also informative and is easier than dealing with telescopes
since it doesn't involve Doppler shift -- you can do it just from the
transforms as given here.

With no further introduction, here it is again.

(Pick holes, if you will -- I'm always willing to learn...)

================================================== =================

The problem as stated was "two twins fly to Earth, each from a distance
of 1 ly, at a velocity of 0.75c. What happens?"

I worked this one through. There's no contradiction. It is, however,
fiendishly confusing.

Typing equations in Ascii is not a lot of fun so I'm going to try to
keep this brief. (Ha, ha.)

I'll try to describe what happens in English, with limited use of
equations. First, I made a few simplifications of the original
statement of the problem to keep the math tractable:

-- Set c=1
-- All dimensions are the same -- time and distance are both measured
as lengths (conversion factor = c, which we set equal to 1).
-- I used a velocity of 1/2 rather than 0.75
-- I assumed that A starts at location -1, B starts at location +1,
and there's an observer in the middle at location 0 who remains
stationary in the initial inertial frame (before anybody
accelerated). Call him "O".

Call the inertial frame they all started in the "base frame".

Then at time 0 in the base frame, an alarm goes off in each spaceship
and at the origin, telling A, B, and O it's time to start. A and B
immediately accelerate to +1/2 and -1/2 respectively.

Gamma and beta are as usual: beta=velocity, gamma=1/sqrt(1-beta^2).
Let's start looking at things from the point of view of 'O'.

In this frame, A is moving at beta=1/2, gamma=2/sqrt(3). The Lorentz
transform for frame(O) -> frame(A) looks something like this, with a
unit width font and a 1-dimensional world (y and z components are
dropped out):

| 2/sqrt(3) -1/sqrt(3) |
| -1/sqrt(3) 2/sqrt(3) |

To get from frame(O) -> frame(B) it looks like this:

| 2/sqrt(3) 1/sqrt(3) |
| 1/sqrt(3) 2/sqrt(3) |

And the transform from A --> O is the same as the one from O --> B
(just reverse the velocity, of course).

Since A starts at X=(0,-1) in the base frame, we transform that to A's
(new) frame and we see that X' = (1/sqrt(3), -2/sqrt(3)).

The distance to the origin (X'=0) went down, 'cause it was contracted,
so it's just 2/sqrt(3) in A's frame. BUT the TIME changed, too --
it's 1/sqrt(3), _not_ 0. This is where the confusing part starts, and
it's worth a few words.

No time passed for A, and his clock still reads 0. However, if, after
accelerating, he _again_ synchronizes his clock with someone located
at X=0, the result won't be the same -- as far as he's concerned, O
and B have both now got clocks that read incorrectly. In simple
terms, he needs to use light-speed signals the sync up his clocks, and
the SOL delay messes up the operation. Since A's clock hasn't
actually changed, and still reads 0, we're going to just subtract
1/sqrt(3) from all times in A's frame from now on. (This makes things
a little messier and is somewhat error prone, unfortunately.)

Anyway, now we've got A with a clock that reads 0, and an apparent
distance to the origin of 2/sqrt(3). He's coming toward O at v=1/2,
so at time t=2 in O's frame, A arrives. In O's frame, the coordinates
are (2,0). We transform that to A's frame, and get
(4/sqrt(3),-2/sqrt(3)) -- BUT we need to subtract 1/sqrt(3) to get A's
actual clock reading. So A's clock must read 3/sqrt(3) = sqrt(3).
That's ~ 1.7, which is rather less than the 2 units that elapsed for
O. A aged less than O.

The behavior of B, from the point of view of O, is identical,
including the subtraction of 1/sqrt(3) to zero his clock. So, as far
as O can see, B also ages by 3/sqrt(3). Since the math is the same
I'm not including it here.

Now, the interesting part: We work it out again from A's point of
view. In A's frame, beta(O) = -1/2, gamma(O) = 2/sqrt(3), and the
transform from A to O is the same as the one from O to B. But we also
need the transform from A --> B, and for that we need beta(B) in A's
frame -- beta'(B), if you will.

B's velocity vector in B's rest frame is (1,0) (moving through time at
rate 1, stationary in space). We can transform that to O's frame, and
we get (2/sqrt(3),-1/sqrt(3)) (this is the 4-velocity with two terms
dropped out, of course). We can now transform that from O's frame to
A's frame using the transform we worked out back at the top, and we
get (5/3,-4/3). The "time" coordinate is just gamma, so gamma(B) in
A's frame is 5/3. The "x" coordinate is gamma*beta, so we divide it
by gamma and see that beta(B) in A's frame must be -4/5. If we plug
that back into the formula for gamma it works out to 5/3 again (which
is a relief): 1/sqrt(1-16/25) = 1/sqrt(9/25) = 5/3.

So the Lorentz transform to get from A's frame to B's frame is

| 5/3 4/3 |
| 4/3 5/3 |

Now, let's go back and figure out how long it takes A to get to the
origin; then we'll look at B again.

A finished accelerating at coordinates (0, -2/sqrt(3)) in his new
frame, including the clock correction. Now, we need to figure out
where the origin is in A's frame. It's _not_ 1 unit away, because
we're measuring the distance to travel in _O_'s frame (and that's the
fundamental asymmetry in this problem, by the way). Looking at O's
frame from A's frame, all distances are contracted, and the length A
needs to travel is 1/gamma = sqrt(3)/2, which is about 0.87.

Now, A sees O approaching at beta=-1/2, so the elapsed time is
(sqrt(3)/2)/(-beta) = sqrt(3). Let's look to see what's happening at
time sqrt(3) in O's frame.

We need to add back 1/sqrt(3) to A's clock and then transform it.
With the 1/sqrt(3) added in, A's coordinates at that point are
(4/sqrt(3),-2/sqrt(3)) (he's stationary in his own rest frame, of
course, so he's still sitting at -2/sqrt(3)). That transforms to
(2,0) in O's frame. Sigh of relief - A and O arrive at the same event
at the same time. Elapsed time for A = sqrt(3), just as we found
before, and elapsed time for O is 2, just as we found before.

Now, let's look at B from A's frame. In the base frame, B's starting
coordinates were (0,1). Transformed to A's frame, they're
(-1/sqrt(3),2/sqrt(3)) ... but with the clock correction of 1/sqrt(3)
that gives us -2/sqrt(3). It seems that B got a head start -- he
started his engine a EARLY from A's point of view. THIS IS IMPORTANT.
The "simultaneous" events in the base frame were NOT simultaneous in
the moving frames of A and B -- and that's where "intuition" totally
jumps the tracks.

If we transform that directly to B's frame using the matrix above
(after adding back that pesky 1/sqrt(3)), we get
(1/sqrt(3),2/sqrt(3)), which we're pleased to see agrees with earlier
results. Of course, we need to subtract 1/sqrt(3) from B's clock, too,
after which we see that B starts at time 0 by _his_ clock.

Now, at time 3/sqrt(3) in A's frame, we find
(3/sqrt(3)+1/sqrt(3),-2/sqrt(3)) maps to (4/sqrt(3),2/sqrt(3)) in B's
frame. Since 2/sqrt(3) is where B's been sitting still in his own
rest frame, A and B do indeed meet at that point. When we subtract
the clock correction of 1/sqrt(3) from B's time we see that his clock
reads 3/sqrt(3) = sqrt(3) -- the same as A's, and the same as what O
predicted. The twins aged the same amount.

Whew.

Again, the problem throughout is that once two frames are moving
relative to each other, they _cannot_ have synchronized clocks -- each
one sees that the other's clocks are skewed progressively with
distance. So, the result works out mathematically, it's supported
experimentally, but it doesn't make a whole lot of sense intuitively.

If you hate the clock corrections -- well, that's the point. The
skewed clocks are what make the math work, and they're what make the
whole thing so confusing.

=========================
To email directly replace nospam with foobox


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  #16  
Old 10-30-2003, 01:58 PM
dlzc@aol.com \(formerly\)
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Default Time dilation - No Need to Do the Math

Dear Stephen Bint:

"Stephen Bint" <[Only registered users see links. ]> wrote in message
news:3fa098d6$0$103$[Only registered users see links. ].. .
moving
velocity

I am assuming I am stationary...

Remember the title of the thread? Since I am disallowed from doing the
math, I can only answer approximately. A little less than 10kph.
Something around the 12th or 18th decimal place.

Relativity only shows up when you move *very* fast, or when you can measure
very precisely. Otherwise Newton (or another of history's geniuses) might
have spotted it.

David A. Smith


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  #17  
Old 10-30-2003, 02:26 PM
sal
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Default Time dilation - No Need to Do the Math

I just posted a long and slightly OT response to this consisting of the
math for the "symetric twin" paradox. Since this discussion is really
just the earlier twin paradox restated, the same analysis applies to both.

However, there's one specific point which should be addressed, and which
Steve Bint should try to grok. Until you get this you'll never see why
the math works out, let alone understand what "really" happens.

Stephen Bint wrote:

When two frames are moving relative to each other, their clocks can NOT
be in sync. What this means, in confusing English, is that when I move
through SPACE in a frame which is moving relative to me, I am also
moving through TIME in that frame! The X axes are not aligned --
they're skewed.

So, the time rates are slowed symetrically ... but the rate of time
change with X position is mirror-imaged between the frames.

If this makes no sense to you (and I expect it does) then you should
really try to work through the math to see why it's asserted as part of
SR. The math, though messy, is _just_ linear algebra -- nothing more.

--
To email me directly, take out nospam and put back foobox.

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  #18  
Old 10-30-2003, 02:33 PM
Harry
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Default Time dilation - No Need to Do the Math


"Dirk Van de moortel" <[Only registered users see links. ]> wrote
in message news:3fa0df79$[Only registered users see links. ].hp.com...
news:bnqmjp$3q7$[Only registered users see links. ]...
news:3fa0893c$0$122$[Only registered users see links. ].. .
it is

What was his old name, do you think?

Harald


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  #19  
Old 10-30-2003, 02:34 PM
Harry
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Default Time dilation - No Need to Do the Math

See below.

"Stephen Bint" <[Only registered users see links. ]> wrote in message
news:3fa07dad$0$118$[Only registered users see links. ].. .

I agree, the faqs can be improved for sure.
Acceleration certainly isn't the official explanation; it's merely an
indicator of asymmetry.
(But you could have noticed, if you read it carefully: "and" instead of
"therefore"...)

SNIP

that

But everyone sees that!

of

True.

in

Acceleration of the ref. frame of observation is not part of SRT, and
therefore not included in the equations either.
However, it is acknowledged that acceleration has no relativistic effect on
the accelerated clock... and then it gets tricky! ;-)
I may post another time more about that point, which is one that made me
choose LET instead of SRT as standard theory.
In case you are really new, LET is the abbreviation used in this group for
Lorentz' interpretation of his transformations and SRT, which differed from
that of Einstein. Nowadays only Einstein's interpretation is taught at
school. No need to tell more, you can find many hours of reading in the
archives... but the same is true for the Twin paradox!

is,
telling
in

Wrong, he did, in 1918; it was -as he put it himself- his clever way out of
the Twin problem (in German, sorry).

will
(predicted

Really? Make sure that you know which acceleration has what effect on which
clock and according to who!

am
return,
reason

Don't overestimate yourself!

Harald


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  #20  
Old 10-30-2003, 02:57 PM
Paul Cardinale
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Default Time dilation - No Need to Do the Math

"Stephen Bint" <[Only registered users see links. ]> wrote in message news:<3fa07dad$0$118$[Only registered users see links. ]> ...

[crap snipped]

I see that you are both a jackass and an idiot.
You believe that because you are too stupid/lazy/arogant to learn the
math, that all you need to do is wave your arms and try to apply your
own ignorant comic-book misconception of relativity. You are a
clueless ineducable blubbering baffoon.

Paul Cardinale
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