What is the physical interpretation of four-velocity in special relativity?

[snoopies622]

This is a bit of a follow-up to my previous question, and may be related to what granpa was just asking. How should one physically interpret the four-velocity vector? For example, if a rocket ship moves past me in the positive x direction with velocity = 0.6c, then (if my math is correct) gamma = 1.25 and the four-velocity vector looks like

< 1.25c, 0.75c, 0, 0 >.

I know that this has a length of c if one uses the Minkowski metric, but since neither I nor the passenger of the rocket ship measure his speed as 0.75c (or 1.25c for that matter) how is this useful or even meaningful?

 

[George Jones]

Think of your spatial coordinate system as a lattice of metre sticks that is at rest with respect to you, and suppose that the ship travels along one of the axes of symmetry of the lattice. Then, 0.75c is the speed at which the ship moves through the lattice with respect to the ship's time, i.e., the number of vertices of the lattice per second that the ship sees whiz by.

 

[yuiop]

Think of your spatial coordinate system as a lattice of metre sticks that is at rest with respect to you, and suppose that the ship travels along one of the axes of symmetry of the lattice. Then, 0.75c is the speed at which the ship moves through the lattice with respect to the ship's time, i.e., the number of vertices of the lattice per second that the ship sees whiz by.

The trouble is that the ship's pilot measures each of the meter sticks to be 0.8 meters each due to length contraction and from that point of view the ship's pilot also measures his own velocity relative to the grid to be 0.6c

What in effect is happening is that four velocity is measured in terms of distance measured by an observer at rest with respect to the grid and time measured by an observer at rest with the ship.

The usefulness of this "reference frame mixing" is that momentum can be expressed as mV where V is the four velocity rather than as mv/sqrt(1-v^2/c^2) where v is the 3 velocity and this avoids all those awkward questions about relativistic mass.

Similarly four acceleration allows force to be expressed in the form f=ma where a is the 4 acceleration and m is the rest mass.

The four velocity of a particle moving at the speed of light is infinite because the proper time is zero and this provides a useful way of explaining why a particle with rest mass can not move at the speed of light without invoking the relativistic mass concept.

Measuring velocity using proper time and improper distance is sometime known as "rapidity" and rapidities can be added and subtracted in the normal way without using relativistic velocity addition equations.

For example if the 3 velocity of a ship is 0.8c relative to observer A then the rapidity is 1.333. If the ship now fires a a missile at 0.8c relative to the ship, which in turn fires a bullet at 0.8c relative to the missile, then the rapity of the bullet relative to observer A is 1.333+1.333+1.333 = 4.0 which is still a long way short of a rapidity of infinity which the rapidity the bullet requires to be moving at the speed of light.

This is a bit of a follow-up to my previous question, and may be related to what granpa was just asking. How should one physically interpret the four-velocity vector? For example, if a rocket ship moves past me in the positive x direction with velocity = 0.6c, then (if my math is correct) gamma = 1.25 and the four-velocity vector looks like

< 1.25c, 0.75c, 0, 0 >.

I know that this has a length of c if one uses the Minkowski metric, but since neither I nor the passenger of the rocket ship measure his speed as 0.75c (or 1.25c for that matter) how is this useful or even meaningful?

The 0.75c in your example is the rapidity that I mentioned above. The magnitude of the four velocity V is the norm of the four vector where

V = \sqrt{ (ct)^2-(x/t)^2-(y/t)^2-(z/t)^2} = \sqrt{1.25^2 - 0.75^2 -0 -0} = \sqrt{ 1.5625-0.5625} = 1

where t is the proper time of the moving ship and distances x, y and z are measured by an observer stationary with respect to the grid.

The four velocity is always equal to c in any reference frame and so must by definition be an invariant. This makes switching reference frames very easy.

The rest mass multiplied by the four velocity gives the four momentum. Since four velocity is invariant and rest mass is invariant then four momentum must also be an invariant and be conserved when switching reference frames.

Four vector type calculations, using proper time have the advantage that the proper time of the moving particle is constant from any reference frame and so that also makes switching reference frames very easy.

 

[granpa]

whats the difference between proper velocity and rapidity?

 

[yuiop]

whats the difference between proper velocity and rapidity?

Technically, proper velocity is always zero. Proper velocity is measurement of an observers own velocity relative to his own reference frame. In relativity, any inertial observer usually considers himself to be stationary.

Rapidity of a ship relative to some observer would be the distance traveled by the ship relative to the observer, divided by the time measured by a clock on board the ship (proper time of the ship).

Normal relative velocity of a ship relative to some observer would be the distance traveled by the ship relative to the observer, divided by the time measured by the observers own clock.

Hope that helps.

 

[granpa]

I'm very confused.

http://en.wikipedia.org/wiki/Proper_velocity

Proper-velocity, the distance traveled per unit time elapsed on the clocks of a traveling object, equals coordinate velocity at low speeds. At any speed it equals momentum per unit mass, and it therefore has no upper limit. It is one of three related derivatives in special relativity (coordinate velocity v=dx/dt, proper-velocity w=dx/dτ, and Lorentz factor γ=dt/dτ) that describe an object's rate of travel. Each of these is also simply related to a traveling object's hyperbolic velocity angle or rapidity η.the article on rapidity is tiny and just talks about hyperbolic angles or some such thing.

http://en.wikipedia.org/wiki/Lorentz_factor#Rapidity

Note that if tanh r = β, then γ = cosh r. Here the hyperbolic angle r is known as the rapidity

 

[Fredrik]

Technically, proper velocity is always zero. Proper velocity is measurement of an observers own velocity relative to his own reference frame. In relativity, any inertial observer usually considers himself to be stationary.

There's another definition of "proper velocity". Check out the Wikipedia article. When someone mentioned it in this forum a few months ago, I had never heard of it. I think most books don't use that concept at all.

Edit: I deleted the rest of the stuff I wrote because I need to think about it some more.

 

[granpa]

Technically, proper velocity is always zero. Proper velocity is measurement of an observers own velocity relative to his own reference frame. In relativity, any inertial observer usually considers himself to be stationary..

exactly what I thought when I first heard the term.

 

[Fredrik]

...three related derivatives in special relativity (coordinate velocity v=dx/dt, proper-velocity w=dx/dτ, and Lorentz factor γ=dt/dτ) that describe an object's rate of travel. Each of these is also simply related to a traveling object's hyperbolic velocity angle or rapidity η.

The rapidity \phi can be defined by \tanh\phi=v. This implies that \sinh\phi=\gamma v and \cosh\phi=\gamma (although it takes some work to prove it). The definition of proper time \tau implies that

\frac{dt}{d\tau}=\gamma=\cosh\phi

and this combined with the chain rule tells us that

\frac{dx}{d\tau}=\frac{dt}{d\tau} \frac{dx}{dt}=\gamma v= \sinh\phi

These are two of the three derivatives you mentioned. The third is just

\frac{dx}{dt}=v=\tanh\phi

 

[granpa]

The rapidity \phi can be defined by \tanh\phi=v. This implies that \sinh\phi=\gamma v and \cosh\phi=\gamma (although it takes some work to prove it). The definition of proper time \tau implies that

\frac{dt}{d\tau}=\gamma=\cosh\phi

and this combined with the chain rule tells us that

\frac{dx}{d\tau}=\frac{dt}{d\tau} \frac{dx}{dt}=\gamma v= \sinh\phi

These are two of the three derivatives you mentioned. The third is just

\frac{dx}{dt}=v=\tanh\phi

is that a way of saying that the rapidity is the magnitude of the spatial component of the four velocity (as per the second part of post 3)? or would that just be proper velocity?

 

[Fredrik]

is that a way of saying that the rapidity is the magnitude of the spatial component of the four velocity (as per the second part of post 3)? or would that just be proper velocity?

According to the Wikipedia article, the proper velocity is the 3-vector you get if you take the four-velocity dx^\mu/d\tau and throw away the 0 component.

The rapidity is definitely not the magnitude of anything familiar.

 

[granpa]

This is a bit of a follow-up to my previous question, and may be related to what granpa was just asking. How should one physically interpret the four-velocity vector? For example, if a rocket ship moves past me in the positive x direction with velocity = 0.6c, then (if my math is correct) gamma = 1.25 and the four-velocity vector looks like

< 1.25c, 0.75c, 0, 0 >.

I know that this has a length of c if one uses the Minkowski metric, but since neither I nor the passenger of the rocket ship measure his speed as 0.75c (or 1.25c for that matter) how is this useful or even meaningful?

suppose you are on a rocket ship with proper acceleration of one g for ten years. seems to me that you would conclude that you were moving at ~10c. that is your proper velocity.

 

[granpa]

According to the Wikipedia article, the proper velocity is the 3-vector you get if you take the four-velocity dx^\mu/d\tau and throw away the 0 component.

The rapidity is definitely not the magnitude of anything familiar.

ok. I think I will stick to proper velocity for now and come back to rapidity later. I have my hands full as it is.

so rapidity has to do with rotation and the unit circle in complex/minkowski space. this sort of makes sense since four velocity always has a magnitude of c.

 

[snoopies622]

Thanks all. Kev you were especially helpful. If rest mass times four-velocity is conserved in collisions (as mv is in Newton's mechanics) than I can see the usefulness of this vector. On the other hand from what I have read I think what you were referring to as the rapidity is actually the hyperbolic sine of the rapidity, which is not additive. Someone correct me if I am wrong.

 

[yuiop]

I'm very confused.

Sorry, I think I was the source of your confusion, shooting fast and loose from dim memory and ended up peppering my feet with (yet more) holes.

Thanks all. Kev you were especially helpful. If rest mass times four-velocity is conserved in collisions (as mv is in Newton's mechanics) than I can see the usefulness of this vector. On the other hand from what I have read I think what you were referring to as the rapidity is actually the hyperbolic sine of the rapidity, which is not additive. Someone correct me if I am wrong.

Yes, I appear to have the proper velocity and rapidity definitions mixed up. I will have to come back to this when I have it cleared up a bit in my head. For now it seems that my discription of rapidity was actually a better description of what is known as proper velocity. Usually all observers agree on proper measurements like proper time and proper distance. The only way I can see that all observer agree on proper velocity is if they define ONE GRID that is the preferred reference frame where all distances used in velocity calculations by any observer are defined only by an observer at rest with that unique grid.

 

[granpa]

ok. I see the difference between proper velocity ond rapidity but now I'm confused about the difference between rapidity and four velocity.

according to the post below I would conclude that rapidity is a vector considsting of 2 parts. one equal to the proper velocity and the other equal to gamma. isn't that basically what four velocity is? (I'm thinking one dimensionally here. maybe that's the trouble)

http://en.wikipedia.org/wiki/Hyperbolic_function

 

[granpa]

nevermind. rapidity is an angle not a point on the circle.

its late. I'm not thinking clearly.

 

[yuiop]

Ok, the part I dimmly remembered about rapidity being simply additive seems to be correct.

This is quote from Baez::

"Sometimes physicists find it more convenient to talk about the rapidity r, which is defined by the relation v = c tanh(r/c)
The hyperbolic tangent function tanh maps the real line from minus infinity to plus infinity onto the interval -1 to +1. So while velocity v can only vary between -c and c, the rapidity r varies over all real values. At small speeds rapidity and velocity are approximately equal. If s is also the rapidity corresponding to velocity u then the combined rapidity t is given by simple addition t = r + s"

See http://www.math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

 

[granpa]

http://en.wikipedia.org/wiki/Proper_velocity

proper velocity also extends the Newtonian form of momentum as mass times velocity to high speeds without a need for relativistic mass[8]

 

[granpa]

Sorry, I think I was the source of your confusion, shooting fast and loose from dim memory and ended up peppering my feet with (yet more) holes.
Yes, I appear to have the proper velocity and rapidity definitions mixed up. I will have to come back to this when I have it cleared up a bit in my head. For now it seems that my discription of rapidity was actually a better description of what is known as proper velocity. Usually all observers agree on proper measurements like proper time and proper distance. The only way I can see that all observer agree on proper velocity is if they define ONE GRID that is the preferred reference frame where all distances used in velocity calculations by any observer are defined only by an observer at rest with that unique grid.

well it seems to me that proper velocity SHOULD be additive. if proper acceleration is in fact the acceleration experienced by the object accelerated (which I haven't seen any proof of) then all observers MUST agree on it. if proper velocity is the integral of proper acceleration then I don't see how it can fail to be additive. all observers should always agree on any value of delta proper velocity.

 

[matheinste]

Hello granpa.

Don't forget that when you integrate a function you have a constant term in the result which in this case is determined by the physical situation. i suspect that in the case of acceleration, when itegrated this contains a constant term which is dependent on the initial velocity and therefore frame dependent. That is a rough description from a non mathematician. Perhaps someone could put it more exactly.i think your problem lies along those lines.

Matheinste

 

[Dale]

rapidity is an angle not a point on the circle.

[nitpick]You mean "point on the hyperboloid"[/nitpick]

By the way, the concept of four-velocity itself can be problematic, since it doesn't really work for light (no proper time so no derivative wrt proper time). However, the four-momentum is well-defined for light. Also, four-velocities are not additive but four-momenta are. When at all possible it is better to work with four-momenta rather than four-velocities since it seems more fundamental and more general. For massive particles then the four-velocity is simply p/|p| where p is the four-momentum and |p| is the rest mass. Essentially it is the unit-four-vector in the "direction" of the four-momentum.

 

[granpa]

but, but, but... four momentum is four velocity times the rest mass which is constant.

 

[granpa]

so what is the velocity addition law for proper velocity?

 

[granpa]

http://Newton.umsl.edu/~run/traveler.html

, if one adds proper velocities w' = gamma' v' and w = gamma v to get relative proper velocity w'', one finds simply that the coordinate velocity factors add while the gamma-factors multiply,

w''=(gamma'*gamma)(v'+v)

 

[Dale]

but, but, but... four momentum is four velocity times the rest mass which is constant.

Yes, you are correct. We are saying the same thing. It might help to write the expressions. I said:
v = p/|p|

And you said:
p = |p| v

Which are mathematically equivalent.

 

[snoopies622]

So if I understand this correctly...

Suppose there are two balls and a compressed spring between them, each with a rest mass of m_0. The four-momenta are

&lt;m_0 c, 0, 0, 0&gt; \\ &lt;m_0 c, 0, 0, 0&gt; \\ &lt;m_0 c, 0, 0, 0&gt;.

After the spring is released, the balls fly off in opposite directions each with speed v. Since four-momentum is conserved, we now have

&lt;m_0 c \gamma, -m_0 v \gamma, 0, 0&gt; \\ &lt;m_0 c (3-2\gamma), 0, 0, 0&gt; \\ &lt;m_0 c \gamma, m_0 v \gamma, 0, 0&gt; where \gamma = (1-v^2/c^2)^{-1/2}

and the spring now weighs m_0 g (3-2 \gamma) instead of m_0 g.

Is this correct?

 

[Dale]

Yes, a compressed spring has slightly less mass than an uncompressed one.

 

[granpa]

but, but, but... four momentum is four velocity times the rest mass which is constant.

Yes, you are correct. We are saying the same thing. It might help to write the expressions. I said:
v = p/|p|

And you said:
p = |p| v

Which are mathematically equivalent.

but you said 'four-velocities are not additive but four-momenta are'.

 

[Dale]

but you said 'four-velocities are not additive but four-momenta are'.

That is correct. The expression x/|x| describes the unit vector in the direction of x. So v = p/|p| means that v is the unit vector in the direction of p. Unit vectors are never additive because the sum of two unit vectors is not in general a unit vector.