Adding Velocities in Relativity

Date: 07/12/99 at 11:05:34
From: Jim Bigari
Subject: Relativity

Traditional math says that if a train is moving 60 mph and a man on 
the train is walking 5mph, someone standing still would say that the 
man on the train is moving 65 mph. However, according to Einstien's 
Theory of relativity, that is a false statement. 

But let's say that the 60+5 = 65 thoery is true. Now let's say the 
train is moving at the speed of light and that the same man is walking 
on that same train at 5 mph. According to pre-relativity math that man 
would be moving at c+5 mph, which is supposed to be impossible. 
Therefore, from a stationary standpoint the man on the train must slow 
down. 

I would be very grateful if you were to send me an equation that one 
could use to calculate the speed of the man on the train (relative to 
the train's floor, not the ground outside). 

An infinite amount of thanks go out to you for such a great Web site 
and the information you provide.

Date: 07/12/99 at 17:25:00
From: Doctor Rick
Subject: Re: Relativity

Hi, Jim, thanks for your question!

Questions like this can be challenging to think about, and even to
state, because in our everyday world (with velocities much less than 
the speed of light) we don't need to be so precise about which 
"reference frame" we mean. Let me restate what I think you are 
asking, and then I'll answer it.

A train is moving at 60 mph as measured by an observer on the ground 
outside - we will say "60 mph in the ground reference frame." A man is 
walking on the train at 5 mph as measured by an observer standing on 
the train ("in the train reference frame"). Then the man walking on 
the train is moving 65 mph in the ground reference frame. That's true 
under non-relativistic ("Newtonian") mechanics. It is also true under 
relativity, as nearly as we could measure, because the train is moving 
at so much less than the speed of light. (Relativity does hold at 
"ordinary" speeds; it's just that the results it gives are very, very 
nearly the same as what we are used to.)

Now, what if the train is moving at nearly the speed of light? (Here 
I must change your question - according to relativity, the train 
couldn't be moving _at_ the speed of light, because it would take an 
infinite amount of energy to get a train up to the speed of light.)

Let's say the train is going 4 mph below the speed of light, c = 
670,680,000 mph. Then if Newtonian mechanics held at these speeds, the 
ground observer would see the man moving at (c-4) + 5 = 670,680,001 
mph. He would be going 1 mph faster than the speed of light. In 
relativistic mechanics, this is not possible. Something must give.

What must give is not the man's walking speed on board the train. As 
far as observers on board the train are concerned, it doesn't matter 
how fast the train is moving; everything behaves normally, including 
how fast the man can walk. (This is a basic principle of relativity 
theory.) But what must give is the way velocities "add." The speed 
seen by the man on the ground is not the sum of the train's speed and 
the man's walking speed; it is somewhat less.

Having modified your question a bit, I can answer it: What is the 
formula for "adding" velocities in relativity? Here is the answer. If

  v' is the velocity of the man in the train frame of reference

  v  is the velocity of the train in the ground frame of reference
   r

  v  is the velocity of the man in the ground frame of reference

  c  is the speed of light

then

         v' + v
               r
  v = --------------
      1 + v' v / c^2
              r

Let's try it out. Suppose the train is traveling at v_r = 0.99c (99% 
of the speed of light), and the man is running at v' = 0.02c. In 
Newtonian mechanics, his speed relative to the ground would be 1.01c. 
But using the equation above, his speed would be

           0.02c + 0.99c
  v = -----------------------
      1 + (0.02c)(0.99c)/c^2)

      1.01c
    = ------ = 0.99039c
      1.0198

The man thus appears from the ground to be moving only 0.00039c faster 
than the train, rather than 0.02c faster. But remember, anyone on the 
train would measure his speed as 0.02c; his speed relative to the floor 
of the train is _not_ reduced.

Relativity physicists describe velocities in terms of a "velocity 
parameter" that does add the way you expect. The relation between the 
velocity and the velocity parameter is something like the relation 
between the slope of a line and the angle it makes with the x axis. 
You can see how velocity is like a slope, if the x axis is time and 
the y axis is distance. But if you draw a second line at the same 
angle to the first line as the first is to the x axis, the slope of 
the second line isn't twice the slope of the first line. Angles add; 
slopes don't. In relativity, velocity parameters add, but velocities 
don't.

As I said, this is a difficult subject to comprehend, but I hope I've
answered your question even if I've raised several more in the 
process.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: Sat, 17 Jul 1999 12:46:34 PDT
From: Jim Bigari
Subject: Re: Relativity

Thank you for your reply! It was EXACTLY what I was looking for. My
friends think that I'm kind of nuts because I enjoy studying math and
physics in my spare time. I'm only 15 and when I come to the teachers
in my school with a question like this they seem a little shocked and I
don't know if they take me seriously.

The graph that you talk about at the end of your letter, is that
Einstein's world line? Anyway, does light bend around a star because of
gravity or because of the fact that the space around the star is bent?
And one more thing, does gravity cause space to bend or does the
bending of space cause gravity? I am sorry if I didn't state those
questions clearly enough but that's the only way that I know how to say
it. Thanks again for everything.

Date: 07/19/99 at 08:52:44
From: Doctor Rick
Subject: Re: Relativity

Hi again, Jim. Let's see what I can say about your new questions, as we
move on from special relativity to general relativity.

Yes, the slopes I described are the slopes of world lines. I didn't
want to get into that too deeply because it would take a book to
explain it properly. The graph behaves a lot like ordinary coordinate
geometry, but it doesn't obey Euclidean geometry - I have heard it
called "Minkowski space" or "Lorentz geometry." (You have probably seen
the names of both those scientists if you have done enough reading to
know about world lines.)

The best answer to both your other questions is that, in general
relativity, gravity _is_ the bending of space - or rather, the bending
of space-time. The cause of both is the presence of mass, such as a
star. The space-time in the vicinity of the star is warped. In the
absence of external forces (excluding gravity, which is _not_ a force
in general relativity), any object will move in a straight line. This
is the same as Newtonian mechanics. However, when space-time is warped,
the definition of a straight line changes. Instead of "straight line,"
we use the term "geodesic," defined as the shortest distance between
two points in the (warped) space.

It's hard to visualize 4-dimensional space at all, let alone _warped_
4-dimensional space. I can't do it! Therefore we explain the concept
with an example using fewer dimensions. The surface of a sphere is a
warped 2-dimensional space. The geodesics on a sphere are "great
circles," like the equator or a line of longitude on the earth -
circles whose center is at the center of the earth.

Any two great circles intersect at two points. Thus, if 2 travelers set
out from the same place in different directions and travel great-circle
routes, they will meet again on the other side of the earth.

This has enough similarity to the bending of light rays to give the
idea. Light rays coming from a star pass on opposite sides of another
star, where the space-time is curved something like the way the surface
of the earth is curved. Like the two travelers, both light rays travel
"straight ahead" (along geodesics), but they meet again, causing us to
say that their paths were bent. We say gravity bent their paths, but
really no force acted on them; they traveled straight through a bent
space-time. The bent space-time _is_ gravity; the bending, or the
gravity, was caused by the mass of the star.

I hope I've made something a little clearer for you. I don't spend
enough time with relativity for it all to be clear even to me. But keep
asking questions and seeking answers. Rest assured that you are not the
only 15-year-old who is interested in math and physics. If your
questions go beyond the range of what your teachers are prepared to
answer, perhaps you could make a contact at a local college. And of
course there's Dr. Math; we'll help as much as we can.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 07/25/99 at 22:40:37
From: Jim Bigari
Subject: Re: Relativity

I am sorry for my ignorance in this section of relativity but I beg you
to be patient with me. I have heard an analogy of bent space-time that
goes like this: Let's say we have a 2D world, like a piece of paper,
and we have 2D people living there. Now let's say that we crunch the
paper into three dimensions (as you crunch the paper it moves up into
the third dimension). Now, the 2D people have been equally crunched so
they do not notice any change at all. But they do notice that there are
now unseen forces acting upon them. For example, they will be walking
on what they see as a flat surface until they reach a crunch in the
paper. Then, as they attempt to walk forward they will feel a force
acting upon them (gravity). Now let's say that it is the same way in
3D. That would mean that bent space-time is just a hill that everything
slides down, but things only slide down because of gravity. It seems
like this whole idea is leading in a circle. Once again, I am sorry if
what I am trying to say is unclear, and I also apologize if this whole
paragraph seems laughably childish, but I don't see a solution.

Thanks again for everything.

Truly curious,
Jimi

Date: 07/26/99 at 08:59:24
From: Doctor Rick
Subject: Re: Relativity

Hi again, Jim, thanks for your questions - they are good ones. There's
no need to apologize for needing to ask questions in order to
understand things. If your teachers or I ever seem impatient, please
understand that we can be frustrated at our own inability to
communicate matters that we may not fully grasp ourselves.

The image of a crumpled sheet of paper isn't a good way to think about
it. A sheet of rubber captures the idea more accurately, but it's still
very easy to misunderstand.

The difference between a piece of paper and a rubber sheet is that the
paper just bends and folds, but it doesn't stretch. Distances and
angles measured along the paper surface remain exactly as they were
before the paper was crumpled. This means that straight lines remain
straight lines, and no "fictitious forces" are introduced.

Look again at the sphere analogy that I mentioned in my first response.
Are you familiar with the map projection problem? In essence, the
problem is that you can't wrap a sphere in paper without stretching the
paper. All map projections have some sort of stretching involved - the
latitude lines aren't parallel, or the longitude lines _are_ parallel
when they shouldn't be, or something like that.

You can take a rubber sheet and stretch it to cover at least a portion
of the sphere. But if you had a square grid marked on the rubber sheet,
the lines would no longer be straight -- they would curve, and lines
that were parallel would no longer be parallel. This is what I
described in my paragraphs last time:

>The surface of a sphere is a warped 2-dimensional space. The geodesics
>on a sphere are "great circles," like the equator or a line of 
>longitude on the earth - circles whose center is at the center of the
>earth.
>
>Any two great circles intersect at two points. Thus, if 2 travelers 
>set out from the same place in different directions and travel great- 
>circle routes, they will meet again on the other side of the earth.
>
>This has enough similarity to the bending of light rays to give the.
>idea Light rays coming from a star pass on opposite sides of another 
>star, where the space-time is curved something like the way the 
>surface of the earth is curved. Like the two travelers, both light 
>rays travel "straight ahead" (along geodesics), but they meet again, 
>causing us to say that their paths were bent. We say gravity bent 
>their paths, but really no force acted on them; they traveled straight 
>through a bent space-time. The bent space-time _is_ gravity; the 
>bending, or the gravity, was caused by the mass of the star.

The lines of longitude (the lines that go from pole to pole) are great
circles. If you start at the North Pole and go straight ahead, you will
follow a line of longitude until you reach the South Pole. If you and I
both set out in straight lines from the North Pole, but in different
directions, we will follow two different lines of longitude. At first
we will get farther and farther apart. But eventually, as we move
around the curvature of the earth, our paths won't diverge quite as
fast. Then as we pass the equator, our paths will actually begin to
converge - we will get closer and closer until we meet at the South
Pole.

If we thought the earth was flat, how would we interpret this? There is
some mysterious force that is pulling us together. That is sort of the
way that gravity works in general relativity. And you can see that this
analogy has nothing to do with the force of gravity acting in 3-D
space; if it did, these travelers would feel themselves being pulled
toward the South Pole, not toward each other.

You may have seen the "rubber sheet" analogy in which planets and stars
are like marbles placed on a thin rubber sheet; their weight makes
little dimples in the surface of the sheet, and these dimples cause
smaller objects placed on the sheet to swerve.

This analogy is confusing because it does invoke gravity in the 3-D
world to produce the dimples, and to keep the smaller objects on the
surface. But that gravity is _not_ what causes objects to swerve; this
is caused simply by the curvature of the sheet. You could freeze the
rubber sheet, so the dimples are frozen in place, then take it into
outer space where there "is no gravity." If you could then find a way
to make objects slide along the surface (magnetize the sheet?), they
would still swerve as they passed through a dimple.

The other confusion caused by all these analogies is that I have only
talked about curved _space_. What is actually curved in general
relativity is _space-time_. To understand it correctly, we would need
to think about what a straight line in space-time means. But I feel
that I am on the edge of what I comprehend myself, so I'd better stop
here for now.

If you'd like to ask more, go ahead. I will have to pull out my copy of
_Gravitation_ by Misner, Thorne, and Wheeler (a huge, graduate-level
book) and see if it can help me explain this better. But for now, I
hope this helps you make some sense of the analogies you have heard.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 07/26/99 at 16:20:13
From: Jim Bigari
Subject: Re: Relativity

Thank you for all of your answers, they really helped me a lot. I now
have some questions on some different terms: There is a classic
equation that is p = mv. When you use that in relativity you write it
as p = gamma mv. The explanation says that "gamma" is the relativistic
factor. How exactly does "gamma" change the equation? Is it a just a
short way of writing another equation that describes some other force?

Also could you send me (or tell me where I could find) the definitions
of terms such as delta and gamma?

Thanks again,
Jimi

Date: 07/27/99 at 08:59:16
From: Doctor Rick
Subject: Re: Relativity

Hi again, Jimi.

In relativity, beta and gamma are used as shorthand for the following
expressions, which appear often:

      beta = v/c

                     1
     gamma = -----------------
             sqrt(1 - (v/c)^2)

Thus, if you know the velocity, you can calculate gamma. When v << c
(velocity is much less than the speed of light),

      beta << 1

     gamma ~= 1 + (1/2)beta^2
           ~= 1

since the second term is much less than the first. Therefore at low
velocities, the momentum p is approximately equal to mv. At high
velocities (approaching the speed of light), on the other hand, when
beta = v/c approaches 1, the denominator in gamma approaches 0, so
gamma increases without limit, and so does momentum p.

Gamma appears in the equation for momentum, not because of any
additional force, but because of time dilation. In relativity, the
passage of time depends on how fast an observer is moving. You have
probably read that time passes more slowly for someone who is moving
than for someone who is watching him move. While one second passes for
the moving person, a longer time passes for the outside observer; this
time is exactly gamma seconds.

Momentum is mass times velocity. Velocity, remember, is the rate of
change of position with respect to time - that is, how many meters an
object moves per second. We need to ask: Whose time do we use in
defining the velocity used in the definition of momentum? It turns out
that the correct way to define momentum is to use the _proper time_ of
the object - that is, the time that you would measure if you were
moving along with the object.

You sit in a "laboratory frame of reference" measuring the distance the
object moves in one second on _your_ clock to find its velocity. But to
compute the momentum you need to multiply the mass by the distance the
object moves in one second of _proper time_. This is longer than one
second of _your_ ("laboratory") time by the factor gamma, so you need
to multiply the mass-times-velocity by gamma to get the momentum.

I don't know what you are referring to when you ask about delta. It is
not used in the way that beta and gamma are used in relativity, as far
as I have seen. But delta is generally used as a symbol for "change."
For instance, if something happens at time t1 and something else
happens at time t2, the time elapsed between the two events is 
delta t = t2 - t1. Velocity is the distance an object moves divided by 
the time it takes, or

     v = delta x / delta t

You can imagine that a complete answer to your questions requires a
book, and a college-level book at that. I hope I've made some sense to
you though I haven't given you all the background you'd need to
understand it fully.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/     

Date: 08/07/99 at 21:05:21
From: Jim Bigari
Subject: Re: Relativity

Hi again!

This may seem like a trivial question, but what is the unit for mass?
And how do you measure something's mass?

Thanks for everything.
Jim Bigari

Date: 08/14/99 at 14:33:06
From: Doctor Rick
Subject: Re: Relativity

Hi again, Jim. 

Mass is measured in kilograms. I imagine you're a bit confused by the
distinction between mass and weight. We customarily say that something
'weighs 20 kg', but when we want to be precise in physics terms, we
must say it 'has a mass of 20 kg', or perhaps that it 'weighs as much
as a mass of 20 kg'. In physics, 'weight' is the force of gravity
pulling on an object, and it should strictly be measured in newtons
(1 N = 1 kg-m-s^-2, or 1 kilogram-meter per second squared). But as
long as we are on the surface of the earth, gravity pulls on two
objects with equal force if they have equal mass. That's why I can say
'this object weighs as much as a 20-kg mass'.

The phrase I just used points the way to how we measure mass. There are
two common types of scale: a spring scale (like most compact bathroom
scales), and a balance scale (the typical doctor's-office scale). The
spring scale measures weight because it compares the force of gravity
with the force exerted by a spring. The balance measures mass because
it compares the force exerted by gravity on the object with the force
of gravity on a standard mass. 

The balance scale itself comes in two types: a 2-pan balance such as
Justice holds in statues, and a beam balance such as the "triple-beam
balance" on which weights are moved along 3 ruled scales. The 2-pan
balance comes with a set of weights (really standard masses), and you
put various combinations of the weights on one pan until they balance
against the object in the other pan. The beam balance has 3 standard
masses attached to it; these masses are moved until the gravitational
forces on them balance the weight of the object by the principles of
levers.

You might still think that the balance compares weights, not masses.
But if you took the balance to the moon, it would indicate exactly the
same mass for the object as it did on earth. The force of gravity on
the object would be less, but the force on each standard mass would
also be less, by exactly the same proportion. Thus, though the balance
requires gravity in order to work, it does not care how strong
gravity is; it really compares the _mass_ of the object against its
standard masses. A spring scale, on the other hand, would indicate a
much lower weight for the object on the moon, because the force of
gravity is less.

I hope this answers your questions. If it raises more questions, go
ahead and ask!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/     

Date: 07/28/99 at 19:49:48
From: Jim Bigari
Subject: Re: Relativity

Hello!
The gamma equation you sent me (as well as other relativistic
equations) has the number 1 in it. Where does this number come from?

Thanks again,
Jimi

Date: 07/29/99 at 17:37:41
From: Doctor Rick
Subject: Re: Relativity

Hi, Jimi.

Where does 1 come from? It's always been there - it's the mother of
all numbers! ;-)

Really, I'm not sure what you're asking. I could go into how the
Lorentz transformation equations can be derived from the invariance of
the speed of light in all reference frames, or from the invariance of
the interval between two events (the space-time equivalent of the
distance between two points). But you'd be better off reading about
this in a book. Maybe it will be enough for me to write gamma in a form
that does not have a 1 in it:

                     1                  c
     gamma = ----------------- = ---------------
             sqrt(1 - (v/c)^2)   sqrt(c^2 - v^2)

If you'd really like to see where gamma comes from, let me know and
I'll try to explain it. But again, a good book will have pictures so it
would be easier to follow.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/     

Date: 07/29/99 at 19:01:44
From: Jim Bigari
Subject: Re: Relativity

Hi,

I have another question. What is the equation for momentum?

Thanks,
Jimi

Date: 07/30/99 at 08:20:27
From: Doctor Rick
Subject: Re: Relativity

Hi again.

You've got me confused. You've already told me the equation for
momentum (assuming we're still in the context of relativity):

     p = gamma m v

where 

                     1
     gamma = -----------------
             sqrt(1 - (v/c)^2)

Is there some other equation involving momentum that you want to know
about? The main facts about momentum remain the same in classical
physics and relativistic physics: the total momentum of a system is
constant if no outside forces act on the system; and in the presence of
a force, the force equals the rate of change of momentum.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/      

Date: 07/30/99 at 17:07:18
From: Jim Bigari
Subject: Re: Relativity

I am extremely sorry! I don't know where my head was!

Anyway, I have another question. They say that if a man was to shoot off at
near-speed-of-light velocities and fly around the universe, when he returned to
earth everyone he knew would have aged considerably, but he would not have. How
is this so, if all effects of the journey are relative to the viewer's
standpoint? Yes, I agree that someone sitting at a "stationary" standpoint would
see the man in the ship move slower, BUT the man in the ship would also see the
people outside move slower. So when he returns his clock and their clock SHOULD
be in tune to each other, shouldn't they?

And one more thing just for clarification. If a ship moved at near c velocities
and someone was looking at it from earth, would the ship appear to slow and
shrink, or just the man inside of the ship? I hope I stated that clearly enough.

Thanks again,
Jim Bigari

Date: 07/31/99 at 08:54:48
From: Doctor Rick
Subject: Re: Relativity

Hi again, Jim.

The last question first, since I can be brief: Anything moving at high
velocity - man, woman, cat, spaceship - would appear shorter to a
stationary observer, and clocks on board - including a heartbeat and
the decay rate of subatomic particles and nuclei - would appear to be
running slowly.

On the first question, I will pass on something I wrote in answer to
another student. He was a little older, but I think you can get
something from this.

>Question:
>Studying relativity for myself, I came across some kind of problem:
>If somebody's time shrinks according to its velocity, can't we say 
>that the people that are staying motionless move at the same velocity 
>relative to the one that is moving? Therefore, isn't their time also
>shrunk?
>
>I'm not sure if it's clear, so I'll try to reformulate it:
>
>A and B have the same age. A leaves the earth for a travel through
>space and therefore goes fast. According to B, the time of A is shrunk
>because he moves fast. But according to A, B is also moving fast, so
>its time should also be "smaller". When they meet at the welcome-back
>party, they should both be younger than the other one?
>Is it right?
>
>Thanks for your help :)
>Florian

The problem is clear and you're not the first person to wonder about it
- it's been called the "clock paradox." I don't know if I can explain
it adequately without giving you a full background in special
relativity, but I'll try.

You just said that A "travels through space and goes fast." But we need
to be more specific. What you have learned about "time dilation" is
usually presented in terms of motion at a constant velocity. We can't
be talking about constant velocity in this case. If A moved at constant
velocity relative to B, he would keep getting farther away, and they
could not meet at a welcome-back party. Therefore A must experience
acceleration - at some point he must turn around and head back.

It is A's acceleration that introduces an asymmetry into the picture. 
B DID travel at uniform velocity the whole time, but A DIDN'T. This
asymmetry is enough reason to say that we need not expect the time-
dilation argument to work in both directions; it is okay for B to age
more than A, and not vice versa.

Would you like a little more detail? Let's get rid of the acceleration
as much as possible - special relativity can handle acceleration, but
it makes the math more complicated. Imagine that A hops on board a
speeding rocket as it passes the earth. After a year (in A's time), the
rocket passes another rocket heading back toward earth, and he hops
from the first rocket to the second. (It would hurt a lot more than
jumping from a speeding train, but let's pretend he could do it!) After
another year in his time frame, A returns to earth and hops off the
speeding rocket, ready for the welcome-home party.

We know what happens from B's point of view. A is traveling the same
speed the whole time (though his direction of travel changes in the
middle). A's time is "shrunk" - the 2 years of A's time is a longer
time for B, so B is older.

What about A's point of view? During the first half of the trip, B was
moving away from him, so B's time was "shrunk" and B should have aged
less than a year at the time that A hops onto the returning rocket.

Suppose there were a network of satellites along the route of the
rocket, all the satellites stationary with respect to the earth. Each
satellite would have a clock synchronized with the earth's clocks. 
Person A could see Earth time as he passed each satellite. He would see 
time dilation: when a clock on his rocket said that a year had passed, 
the Earth-time clocks would say that less than a year had passed.

Likewise on the return trip, B (on earth) would be moving toward A, so B
should age less than a year during the return trip. A should be younger
than B at the party, right?

The key to the paradox is a notion that you might not have heard of. We
hear about length contraction and time dilation, but another effect of
high-speed motion is relativity of simultaneity. If A could see the
times on two of the satellite clocks, one behind him and one ahead of
him the same distance (so the light reaching him from both clocks would
be delayed by the same amount), he would see that the clocks weren't
synchronized. Looking back toward earth, the time would be earlier.
Person A could figure out what a clock on earth must read at that
moment, by extrapolation of the time shift from one clock to the next.
Remember, it's "that moment" in A's reference frame. In the earth's
reference frame (and that of the satellites), the clocks ARE
synchronized; they all read the same time at the same moment.

Now A jumps onto the earth-bound rocket. Suddenly this situation
reverses. Looking toward earth, which is now ahead of him, the earth
time would be LATER rather than earlier. If he figures out what time it
is on earth now, he will find that it is much later than it was just a
moment ago when he was still outbound. During the (very rapid)
acceleration, time on earth made a big jump.

Do you see the implication? B (on earth) was aging more slowly than A
because B was moving relative to A. But now all of a sudden B is older
than A. During the return trip, B again ages more slowly than A, but
not enough to keep B from being older than A when they meet again.

If you're not confused, I'm surprised. I just hope I have given you
some reason to see that time dilation can work, and this isn't really a
paradox.

If you want to look into this further, I have a book called _Spacetime
Physics_ by Edwin Taylor and John Wheeler, published by W. H. Freeman.
It discusses paradoxes like this in great detail and in a fairly
understandable fashion. I don't know if it's still in print, but you
might try to find it.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/      

Date: 07/31/99 at 11:24:42
From: Jim Bigari
Subject: Re: Relativity

Hi,
If a ship traveling at near c velocities appears to slow from a "stationary"
standpoint, and e = mc^2 (so energy and matter are actually one of the same) how
come light can even move? Shouldn't it appear to move EXTREMLY slow or stop
completely? And one more question about your answer to the time paradox: If the
distance traveled from the earth to wherever he stops and back is the same, and
he never slows down or stops, shouldn't the trip back cancel out all time
changes that occurred on the trip out? Once again, I am sorry if I missed
something in your answer, which is probably the case.

Thanks for everything,
Jim Bigari

Date: 07/31/99 at 23:28:12
From: Doctor Rick
Subject: Re: Relativity

Hi, Jim.

A ship traveling at a relativistic velocity does not appear to move
more slowly; rather, _clocks_ on board would appear to move more slowly
than stationary clocks, from the point of view of a stationary
observer. Speed is speed: if the ship is moving at 0.9c relative to
you, that is what you will measure; and to someone on the ship, you
will appear to be moving at 0.9c in the opposite direction.

Actually, light can _only_ move at the speed of light, because it has
no rest mass m. A photon cannot be stopped or even slowed down.
Anything that has rest mass, on the other hand, can't move at the speed
of light because it would take infinite energy to reach that speed. If
you could move at the speed of light (but you can't because you have
non-zero rest mass), time would appear to stop; but you would still be
moving at the speed of light relative to anything that wasn't moving
with you.

Though light has no rest mass, it does have momentum. If you recall the
formula for momentum,

     p = mv / sqrt(1 - (v/c)^2)

and you set v = c and m = 0, you get 0/0, which is indeterminate - you
can't tell what the correct answer is. But by other means it can be
shown that 

     p = E/c

for something moving at the speed of light. The energy is not infinite,
as it would be for anything with non-zero rest mass; in the case of a
photon of light, the energy turns out to be proportional to the
frequency of light, but this is not part of relativity theory - it's
quantum mechanics.

>And one more question about your answer to the time paradox: If the 
>distance traveled from the earth to wherever he stopped and back is 
>the same, and he never slowed down or stopped, shouldn't the trip 
>back cancel out all time changes that occurred on the trip out?

The time dilation experienced on the outward trip of the spaceship is
not canceled by time dilation on the return trip. Time dilation depends
only on the relative speed, not the direction of motion. Therefore time
slows down for the traveler in both directions, so that when he returns
he is younger than the earth-bound person.

Time also slows down for the earth-bound person from the point of view
of the traveler, during those two legs of the trip. As I explained,
it's during the turn-around (or the jump between spaceships, as I
described it) that the traveler will "see" the earth-bound person age
rapidly.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/      

Date: 08/02/99 at 17:26:40
From: Jim Bigari
Subject: Re: Relativity

Can we turn energy into mass? I know we can turn mass into energy
(atomic bombs), and I know it is possible because of e = mc^2, but do
we have the technology?

Thanks again
Jim

Date: 08/02/99 at 18:08:13
From: Doctor Rick
Subject: Re: Relativity

Hi, Jim.

An interesting question! 

As you know, the fact that c^2 is so big means that just a little loss
of mass in the fission of uranium or plutonium, or the fusion of
hydrogen, releases a whole lot of energy. For the same reason, it takes
a whole lot of energy to make just a little mass. That makes the
reverse process pretty hard.

Making mass from energy is one of the functions of a particle
accelerator. But it isn't worthwhile to try to make ordinary matter
this way; instead, scientists use accelerators to try to make new kinds
of matter. For instance, an electron-positron collider like LEP at
CERN,

     http://www.cern.ch/Public/ACCELERATORS/Welcome.html   

accelerates electrons to high energies in a huge circular tunnel. It
accelerates positrons (positively-charged versions of electrons) in the
same tunnel but moving in the opposite direction. When an electron
collides with a positron, the electron and positron annihilate each
other, turning into a pair of photons. Under the right conditions,
these photons (which have no rest mass, only energy) turn into unusual
particles that are much heavier than ordinary particles like protons or
neutrons. These particles don't last very long before they decay into
lighter particles, but their existence can be detected. This is how
CERN made the Z and W particles, which carry the "weak force" just as
photons carry electromagnetic force.

So the answer to your question is yes, the technology does exist to
make matter from energy. But the matter it makes isn't in the form of
molecules or atoms or nuclei. All those things, even nuclei apart from
hydrogen, are very complex, and the technology does not exist to smash
things together at high energy and make _complex_ matter come out.
Smashing things together just isn't very neat.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/      

Date: 08/02/99 at 19:52:38
From: Doctor Ian
Subject: Re: Relativity

Hi Jim,

If you accelerate an object - which requires transferring kinetic
energy to the object - at low speeds, most of the energy goes to
increase the speed of the object; but as the speed approaches the speed
of light, most of the energy goes into increasing the mass of the
object. This has to happen because there is a limit to how fast
something can go, but there is no limit to how much mass something can
have. We do, in fact, have the technology to do this - we do it every
time we accelerate particles to significant fractions of the speed of
light in a particle accelerator.

And when you smash particles together in a particle accelerator, the
energy of collision sometimes goes into creating particles with greater
rest masses than the particles you started with. And of course, any time
you create photons with enough very high frequencies, some of them
split into particle/anti-particle pairs.

So the answer to your questions are 'yes' and 'yes'.

But just as exploding an atomic bomb releases an enormous amount of
energy from a small amount of mass, creating a small amount of mass
would require the input of an enormous amount of energy. So, except as
a way to test scientific theories, it's not clear what would be gained
by doing it.

But then, who expected lasers to be useful for playing music?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/     

Date: 08/02/99 at 20:49:32
From: Jim Bigari
Subject: Re: Relativity

Is there any proof that other dimensions do REALLY exist? Because most
of the newer theories are based on ten dimensions, like the superstring
theory and Kaluza-Klien Theory. One day, are all these theories going
to crumble to the ground because they are based on false notions? I
just got done reading a book by Michio Kaku called _HYPERSPACE_ and it
seems like they throw dimensions around as if they were just a very
minor thing. It seems as if their formulas don't fit nature, then it is
not their formulas and equations that are wrong... IT'S NATURE! I just
thought that was odd, but it is probably because I failed to grasp
something.

Thanks for everything,
Jim Bigari

Date: 08/02/99 at 22:31:27
From: Doctor Ian
Subject: Re: Relativity

Hi Jim,

In science, you can't really 'prove' anything. The best you can hope
for is to find evidence in support of a theory, or evidence that
contradicts the theory. But even when you can't find any contradictory
evidence, you still haven't 'proven' anything, because there might
always be other theories you haven't thought of that would explain
things just as well. 

Basically, once you get past three dimensions, it's impossible to
really visualize what's going on, so you just have to trust that if
you've derived your equations correctly, they'll tell you what's going
on - even if what's going on doesn't seem to 'make sense'. It turns
out that there have been many instances in the history of science where
a scientist found what seemed to be a bizarre mathematical result,
concluded that the result must be true, dreamed up an interpretation
for the result, and discovered that the interpretation was confirmed by
experiments.

One such discovery occurred when Paul Dirac noted that some of his
equations had two sets of solutions: one positive, one negative. He
could have concluded that the equations were wrong, but he thought
about it instead, and predicted the existence of anti-particles - a
prediction that was quickly confirmed, and which opened the way for
all kinds of other theories. Another example was Louis De Broglie's
prediction that particles have associated wavelengths. When he first
proposed the idea in his doctoral thesis, Einstein was the only person
who took it seriously. Kaku quotes Niels Bohr as saying about one of
Pauli's theories: "We are all agreed that your theory is absolutely
crazy. But what divides us is whether your theory is crazy enough." And
he quotes Richard Feynman as saying: "I think it is safe to say that no
one understands quantum mechanics. But do not keep saying to yourself,
if you can possibly avoid it, 'But how can it be like that?'.  Nobody
knows how it can be like that."

Feynman once gave a series of lectures for non-scientists that were
collected into a book called _The Character of Physical Law_, which
contains a lecture called "The Relation of Mathematics to Physics." The
book is widely available and easy to read, and I think that chapter
would address the questions you've asked, and others that you haven't.
By the way, anything written by or about Feynman is definitely worth
reading. His autobiography, _Surely You're Joking, Mr. Feynman_, is a
good place to start. 
 
- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/