The **Dog School of Mathematics** Presents

**10. Addition of Velocities and Increasing Mass.**

In the beginning of these "lessons" I derived the **Galilean Addition-of-Velocities
Formula**. It's the pre-relativity answer to the question:

If I'm running 10 miles per hour down the aisle of the trainThe pre-relativity solution is simply 25 mph + 10 mph = 35 mph. For all practical purposes that's also the relativistic solution at such slow velocities.

and the train is going 25 miles per hour down the track then

how fast am I running when measured against the track?

Let's go back and review the derivation I used but we'll now give it a relativistic criticism:

We were answering the question:

If I am going at velocityHere is the derivation from Chapter 0:vrelative to you and a rocket is going

at velocityurelative to me how fast is the rocket going relative

to you?

If the railroad flatcar is going at velocity v with respect to (WRT) you then after t seconds t will have moved a distance vt away from you. If I am going at velocity u with respect to the flatcar then after the same t seconds I will have moved a distance ut along the flatcar. How far have I moved away from you? Well, the flatcar has moved vt and I have moved an additional ut so the total distance I have moved is ut + vt = (u + v)t.

This can be written as (u + v)t. Now speed is the thing you multiply times time to get the distance. Here we multiply time t by (u + v) to get the distance I have traveled away from you. That means that relative to you I am traveling at the speed

speed of me relative to you = u + v

That was our derivation way back in SR (0). We used the "normal" concepts of space and time (the ones used by Galileo and Newton) in the derivation.

Now we see some problems. The distance ut that the rocket went relative to me is not the same distance away from me when it is measured in your frame. There are at least two (three?) things to consider:

1) The distance labeled ut (measured in my frame) is contractedThus we should NOT expect this formula to hold under the assumptions of Special Relativity. The Relativistic formula for addition of velocities is: where

because of length contraction when it is measured in your

frame.2) The "t" in the formula ut was the elapsed time as measured

with my clocks which area) running slower than your clocks (when measured by you) andb) not synchronized. When I measure the elapsed time I need to

use two different clocks. The clock where the rocket begins

and the clock where the rocket ends the timing. I observe

these two clocks as synchronized but you view them as

not synchronized.

uis my velocity measured in YOUR framevis the rocket's velocity measured in MY framewis the rocket's velocity measured in YOUR framecis the speed of light.

Notice that the top of the formula is simply *u + v* which is
the "old" Galilean formula--the normal way. This is divided by
(*1 + uv/c ^{2}*)--a sort of "fix-up factor". If u and
v are small compared with the speed of light then the fraction

Let's plug some numbers in the formula for speeds comparable to c and see how it behaves.

If I am going at .5 c relative to you and the rocket goes .5c relative to me how fast will it go relative to you? In this case both u and v are .5c and the formula gives

w = (.5c + .5c) / (1 + (.25cAfter we do some "simplifying" we get^{2})/c^{2})

w = .8cNotice that we haven't succeeded in reaching the speed of light.

What if I'm going .5c and I turn on my headlights. How fast will you see the light from my headlight going? (Note: by the second postulate this had better be c or else something's wrong!)

.w = (.5c + c)/(1 + (.5cSo the formula is consistent with the first postulate.^{2})/c^{2}) = (1.5c)/(1.5) = c

There are two consequences that follow from the formula for relativistic addition of velocities.

If you are going less than the speed of light and you speed up from your frame by less than the speed of light then your final speed is still less than the speed of light.1. You can't reach the speed of light by speeding up.

[Let's say that in our "extended language" -- If I are moving withLet x and y be fractions (greater than 0 but less than 1). Then

velocity u < c in your frame and I fire a rocket with velocity

v < c as measured in MY frame then the rocket will have velocity

w < c as measured in your frame.]

the two speeds less than c can be written as xc and yc. If we

write u and v as xc and yc then the formula gives (after a bit of elementary algebra)

w = (xc + yc)/(1 +(xycIf x and y are between 0 and 1 it is a mathematical result that (x + y)/(1 + xy) will also be between 0 and 1. Let's try .99 for both x and y. then^{2})/c^{2}) = (x + y)/(1+xy)

w = (.99c + .99c)/(1 + (.99c .99c)/cNow the fraction 1.98/1.9801 is slightly less than 1 since its numerator is less than its denominator.^{2}) = (1.98c)/(1.9801)

That means that if I begin with velocity less than c as measured in your frame then by speeding up (adding a velocity) by less than c will not leave me going faster than c as measured in your frame. Also, doing this "speed up" over and over again will never get me past the speed of light.

What, in physics, is mass. It is the property of a body to resist acceleration. Newton's second law is2. "Mass Increase"

WhereF = Ma

We can divide by **a** to rewrite the equation as

Note that the larger the mass M the larger the force has to be to produce the same accelerationM = F/a

How does this enter in to relativity?

Let's posit the Standard Rocket. It has a standard mass and has a standard thrust which produces a standard acceleration for a standard time. The point is that all of these rockets are identical. Since they are identical they will all have the same final velocity after their engine is finished firing (measured relative to the inertial frame in which they began).

Now suppose that we fire off one of these rockets. It is traveling through space some speed v. (measured by us) Our assistant in a large spaceship catches up with the rocket and is "drifting" alongside the rocket with another standard rocket. He points the standard rocket in the same direction as the first rocket.

At this point both rockets are traveling with speed v measured in our frame. He ignites his standard rocket and it accelerates away from him until it has a velocity v relative to him. Now under the old Galileo/Newton addition of velocities his rocket would be going at speed 2v as measured by us. However with the relativistic formula for addition of velocities this is not the case.

Under the relativistic formula his rocket is going LESS THAN 2v when measured in our frame. This means that the standard rocket had less acceleration when it was moving with respect to us than when it was at rest with respect to us.

[Firing the engine increased its velocity from v to less than 2v. Thus the change in velocity was less than v. It didn't accelerate as much.]

So we can see that moving objects resist being speeded up more than stationary objects.

[translation: Objects which are moving in our frame resist change in velocity, when that change is measured in our frame, more than objects which are at rest in our frame.]

Since, by Newton's second law, mass is what resists acceleration we can say that the mass of the rocket has increased. When it is moving with velocity v it is more difficult to speed up than when it is at rest.

Of course in the frame of the rocket there is no mass increase. Our assistant observes the rocket speeding up to v (in his frame) just as we observed the first rocket speeding up to v in our frame.

Two of the most sacred laws of physics are

The old expressions for energy and momentum involved the mass M of the objects involved. In relativity these old expressions didn't work any more. Using the old formulas for energy and momentum energy wasn't conserved and neither was momentum. Einstein, in a leap of deep intuition, changed the formulas for momentum and energy so that they would be conserved in relativistic interactions. The result was that the in the new formula for energy the energy of an object was not zero when its velocity was zero. This had been the case with the old formula. In fact with the new formula at zero velocity there was some energy left over which was expressed in the famous equation:1) the conservation of energy -- the proposition that energycan be neither created from nothing nor destroyed onto nothing.

2) The conservation of momentum -- you can't start moving withoutsomething to push against.

Next: 11 Faster Than Light?E = Mc^{2}

Send any comments, criticisms or reactions to: