4-velocity vector!

1. Sep 22, 2005

yukcream

I want to know why the 4-velocity vector of a particle is
u'=dx^i/ds where ds=cdt sqrt{1-v^2/c^2}
not dx^i/dt?

also can anyone show me the devide of the 4-acceleration vector,i.e
a' = d^2x'/ds^2?! I dont konw how to write it back in terms of v,c &x^i !!

In addition why the 4-velocity vector is a dimesnsionless quantity? what does it really mean?

thx for help
yukyuk

2. Sep 22, 2005

robphy

One can show that the quantity $$\frac{dx^i}{dt}$$ doesn't transform correctly under a Lorentz Transformation. Hence, it is not a 4-vector.

The quantity $$u^i=\frac{dx^i}{ds}$$ where $$ds=cdt \sqrt{1-v^2/c^2}$$ does transform correctly.
As you have noticed, this quantity is dimensionless. Thus, $$u^i$$ can be regarded as a certain unit 4-vector, namely, the unit timelike 4-vector tangent to the particle's worldline. This normalization is convenient because it absorbs some pesky factors of c. An important role of this unit timelike 4-vector is that it helps decompose ["write as a vector sum"] tensorial quantities into timelike and spacelike parts ["components"] in the frame of the particle.

The 4-acceleration is $$a^i = \frac{d}{ds}u^i=\frac{d}{ds}\frac{d}{ds} x^i$$. This can also be written as $$a^i=u^a \nabla_a u^i$$. To write this in component form, apply the chain rule to $$a^i =\frac{d}{ds}\frac{d}{ds} x^i$$. For simplicity, you'll want to assume that all of the motion is in the x-direction.

Here are some sites to look at
http://farside.ph.utexas.edu/teaching/em/lectures/node115.html
http://vishnu.mth.uct.ac.za/omei/gr/chap2/node4.html

Last edited by a moderator: Apr 21, 2017
3. Sep 22, 2005

Trilairian

Making it unitless is merely someones convention. Not everyone does that. The displacement four-vector $$dx^{\mu }$$Is a rank one tensor, a four-vector and in fact the way dx transforms is usually taken as the definition for what constitutes a tensor. $$ds=cd\tau$$ and ds is obviously invariant and so in that sence so is $$d\tau$$ and tensor multiplied or divided by an invariant remains a tensor and so defining $$U^{\mu }$$ by $$U^{\mu } = dx^{\mu} /d\tau$$ or dx/ds in your case guarantees that you have defined a four-vector. This is then named the velocity four-vector, or 4-velocity vector in your terms. dx/dt is not a four-vector simply because it does not transform like a tensor. The covariant derivative of a tensor is also a tensor and so $$A^{\mu } = DU^{\mu }/d\tau$$ defines a four-vector called the acceleration four-vector, or in your terms the 4-acceleration vector. For intertial frames in special relativity this can be written $$A^{\mu } = d^{2}x^{\mu }/d\tau ^2$$ To write it in terms of v, $$A^{\mu } = (\gamma d/dt)(\gamma dx^{\mu }/dt)$$
$$A^{\mu } = \gamma (\gamma (d^{2}x^{\mu }/dt^{2}) + (dx^{\mu }/dt)(d\gamma /dv)(dv/dt))$$
$$A^{\mu } = \gamma (\gamma (d^2x^{\mu }/dt^{2}) + (dx^{\mu }/dt)((v/c^{2})(\gamma ^{3}))dv/dt)$$
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }((v/c^{2})dv/dt)$$
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$

(Im choosing $$\mathbf{v}\cdot \mathbf{a}$$ to represent the ordinary three component dot product of coordinate velocity and coordinate acceleration and using v and a to represent those even when I choose to give them a fourth element as indicated by the greek index.)

Last edited: Sep 22, 2005
4. Sep 22, 2005

pervect

Staff Emeritus
ds, usually written as $d \tau$, is the proper time of the particle.

The reason that the 4-velocity is the derivative of distance with respect to proper time is to ensure that the 4-velocity is a 4-vector.

4-vectors must have an invariant Lorentz interval. Taking the derivative of (ct,x,y,z) with respect to coordiante time t does not give an invariant lorentz interval, it gives a coordinate dependent quantity. (x,y,z,t) is a 4-vector because it's Lorentz interval is invariant for all observers. To maintain this Lorentz invariance, we have to differentiate with a Lorentz invariant time. The only time available that fits this bill is the proper time of the particle.

I'm usually used to assuming c=1 when dealing with 4-vectors, so I'll let someone else explain the conventions usually used when c is not assumed to be 1.

Note that the "length", i.e. the Lorentz interval, of the 4-velocity is a constant. This implies that the 4-velocity and the 4-acceleration are always perpendicular.

You've already written down the expression for 4-acceleration

$$\left( \frac{d^2 t}{d \tau^2}, \frac{d^2 x}{d \tau^2}, \frac{d^2 y}{d \tau^2}, \frac{d^2 z}{d \tau^2}\right)$$

(modulo some factors of 'c'), so I'm not sure what your question is. Here you write t,x,y,z as functions of $\tau.$. If instead you have x(t), y(t), z(t), you need to find $\tau(t)$, which you can do from the definition of the Lorentz interval, then you need to change varabiles to find x,y, and z in terms of tau, rather than t.

5. Sep 22, 2005

pmb_phy

yukcream - In special relatity should keep in mind that is must be possible to express all of the mathematical quanties used in laws of nature which does not depend on the form of coordinate system chosen. In order for this to work properly each component must have the same dimension.

That the 4-vecocity is dimensionless is a matter of choice and nothing else.

Pete

Last edited: Sep 22, 2005
6. Sep 22, 2005

yukcream

To pervect:
My problem is I don't know how to write a^i as the vector form, like
{gamma, vgamma} for u^i, the 4-velocity vectors!
where gamma= 1/sqrt[1-v^2/c^2]

To Trilairian:
Thank you very much for showing me the devide step by step! I think the dt there is represent the proper time, right?

One more question, does Trilairian's devide give me the spatial component of the 4-acceleration vextor? How about the temporal component?

P.S: Thanks for all who answered my question!

yukyuk

7. Sep 22, 2005

Trilairian

You're welcome. The $$\tau$$ refers to proper time. The t refers to coordinate time(your coordinate's time).

It gives both. For the temporal component plug in $$\mu = 0$$ and use $$a^{0} = 0$$, and $$v^{0} = c$$. (But take note that it seems your instruction uses c = 1, whereas I make no such demand. Such trivial differences in definitions tend to vary from one author to another.)

Last edited: Sep 22, 2005
8. Sep 22, 2005

pervect

Staff Emeritus
Let us suppose you have a particle.

You plot the coordinates of the particle as a function of it's proper time, $\tau$.

You do this by specifying four functions

$\left( t(\tau), x(\tau), y(\tau), z(\tau) \right)$

Then you can easily compute it's 4-velocity, which is the derivative of the above expression with respect to tau, and it's 4-acceleration, which is the second derivative of the above expression with respect to tau, i.e.

4 velocity $$u^i = \frac{d x^i}{d \tau}$$

4 acceleration $$a^i = \frac{d^2 x^i}{d \tau^2}$$

(x(t), y(t), z(t)), rather than the above 4 functions.

You then have to compute $\tau$ by the relationship

$$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$

(This is assuming a flat metric - if you are doing general relativity, you havea to use the more general expression

$$d\tau^2 = g_{ab} dx^a dx^b$$

where you sum over a=0..3 and b=0..3)

It may be worthwile to re-write the above expression to explicitly solve for tau in a Miknowskian space-time, one with a global Lorentz metric:

$$\tau = \int \sqrt{1 - \left( \frac{dx}{dt} \right)^2 - \left( \frac{dy}{dt} \right)^2 - \left( \frac{dz}{dt} \right)^2 } dt$$

This gives you $\tau(t)$. You then have to invert this expression to find $t(\tau)$. Then you can compute the other expressions by substitution, which are $x(t(\tau)), y(t(\tau), z(t(\tau))$

Fortunately, this does not have to be done very often. Usually, one can specify the a^i directly, then find u^i and x^i, and then (if one is interested), convert them into coordinate form via taking $x(t) = x(t(\tau))$, etc.

9. Oct 6, 2005

yukcream

To Trilairian

I know it is too late to ask you, still hope you will answer this!
why a^o is zero? du^0/dt not equal zero ~ where t is the popertime~
why v^o = c? v^0 actually is 1/{1-(v^2/c^2)}

yukyuk

10. Oct 6, 2005

Mortimer

You may want to have a look at http://www.rfjvanlinden171.freeler.nl/4vectors/index.html (of myself). It doesn't encompass acceleration 4-vectors but gives some general considerations regarding the physical interpretation of Minkowski 4-vector components that may answer your other questions.

Last edited by a moderator: Apr 21, 2017
11. Oct 6, 2005

yukcream

To Mortimer:

Thanks for your article but I hope to read an article more related to 4 acceleration as I get in trouble in writing out the component form of 4- acceleration vector~

yukyuk

12. Oct 6, 2005

robphy

If I am understanding the notation correctly, I believe that a^0=0 and v^0=c corresponds to evaluating those 4-vectors in the instantaneous rest-frame of the object being accelerated. In that frame, v is a purely temporal 4-vector (i.e., the spatial velocity is zero, and so gamma is one), and a is a purely spatial 4-vector.

13. Oct 6, 2005

Trilairian

I am using Uppercase A and U for four-vector acceleration and four-vector velocity. I am using lower case $$a^{\mu }$$ and $$v^{\mu }$$ for coordinate acceleration and coordinate velocity and extending their definitions to include a fourth element in the instances where a greek index is used.
$$v^{\mu } = dx^{\mu }/dt$$
$$a^{\mu } = dv^{\mu }/dt$$
For the fourth element:
$$v^{0} = dx^{0}/dt$$
$$v^{0} = dct/dt$$
$$v^{0} = c$$
$$a^{0} = dv^{0}/dt$$
$$a^{0} = dc/dt$$
$$a^{0} = 0$$

14. Oct 6, 2005

Trilairian

As an example this is how you calculate the time element of the acceleration four-vector.
$$A^{\mu } = \gamma ^{2}(a^{\mu } + \gamma ^{2}v^{\mu }(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{2}(a^{0} + \gamma ^{2}v^{0}(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{2}(0 + \gamma ^{2}c(\mathbf{v}\cdot \mathbf{a}/c^{2}))$$
$$A^{0} = \gamma ^{4}(\mathbf{v}\cdot \mathbf{a}/c)$$

15. Oct 7, 2005

yukcream

what will be the result of $$\mathbf{v}\cdot\mathbd{a}$$?
will it be zero as velcoity and acceleration are orthognal to each other!

16. Oct 8, 2005

Mortimer

Last edited by a moderator: Apr 21, 2017
17. Oct 9, 2005

yukcream

I have read an article it states that the time-component of 4 acceleration vector is zero, as in 4 velocity vector, $$v^{0}=v$$ so $$a^{0}=0$$. But in fact it is not correct! It should be a non zero quality.
yukyuk

18. Oct 9, 2005

Physics Monkey

yuk,

Because the four velocity is orthogonal to the four acceleration, the time component of the four acceleration will be zero in the rest frame of the particle (where the four velocity has only a time component). This is probably what the article meant, but who knows?

19. Oct 11, 2005

Trilairian

They aren't generally orthogonal. You are mixing them up with Four-vector acceleration $$A^{\mu }$$ and four-vector velocity $$U^{\mu }$$. These are not the same things as coordinate acceleration $$a^{i }$$ and coordinate velocity $$v^{i }$$. You need to look more carefully at the notation that I am using. I clearly defined things and the way I did so caps makes a difference.

Last edited: Oct 11, 2005
20. Oct 11, 2005

Trilairian

No you didn't. It said the time component of the coordinate acceleration is zero, not of the four-vector acceleration.
1st $$v^{\mu }$$ was not the four-vector velocity, $$U^{\mu }$$ was.
2nd that isn't what was said. What was said wat the time component of the coordinate velocity was c. $$v^{0} = c$$
No, that is correct. You're confusing it with $$A^{0}$$. I clearly defined my notation. Feel free to reread it.