# Special relativity and speed of body

1. Dec 15, 2009

### kapv89

1. The problem statement, all variables and given/known data

a particle is under constant 4 acceleration (given of const magnitude(alpha) and direction), the particle starts from rest in a frame O, i have to find the speed of body after time t in O

2. Relevant equations

well can't give any very important equation, i guess this is pretty basic special relativity, but i am having problem solving this question,

3. The attempt at a solution

i tried to make a relation b/w the magnitude of four acceleration and the speed, but the integration is pretty lengthy and complicated, i feel their must be an easier method

the relation i got was ((gamma)^4 * (a)^2 * (1+ (v^2)(gamma^2)))= (alpha^2)

a= dv/dt v is the speed in frame O

Last edited: Dec 15, 2009
2. Dec 15, 2009

### jdwood983

I think you are making this a little too hard. The 4-acceleration is given by (using differential geometry notation):

$$a^\mu=\left(\begin{array}{c}\gamma\dot{\gamma}c \\ \gamma\dot{\gamma}\mathbf{v}+\gamma^2\dot{\mathbf{v}}\end{array}\right)=\left(\begin{array}{c}\gamma\dot{\gamma}c \\ \gamma\dot{\gamma}\mathbf{v}+\gamma^2\mathbf{a}\end{array}\right)$$

So how can you relate the four-vector acceleration above to the magnitude $\alpha$ given in the problem?

NOTE: $\gamma$ is the normal Lorentz factor, $\mathbf{v}$ is the normal 3-velocity, and $\mathbf{a}$ is the normal 3-acceleration.

3. Dec 16, 2009

### kapv89

i took the A.A= (alpha)^2 where A is the four acceleration and the value of A which i took was same as that u have written.

also, i took d(gamma)/dt = ((gamma)^3)(v)(dv/dt) [which is obvious]

also if i wasn't clear earlier, A.A = (alpha)^2 was the exact statement made in the book, i.e 'schutz: first course in gtr' chapter 2 problem 19 and i am working in units with c=1

Last edited: Dec 16, 2009
4. Dec 16, 2009

### jdwood983

So then just use the definition of Equation (2.24) from what I have written above, recalling the fact that $(\gamma\dot{\gamma}\mathbf{v}+\gamma^2\mathbf{a})^2=(a^1)^2+(a^2)^2+(a^3)^2$.

5. Dec 16, 2009

### kapv89

thnx, got it, was pretty easy though, no need of squaring and all.

6. Dec 16, 2009

### jdwood983

if you don't mind my asking, how did you accomplish this without squaring?

7. Dec 16, 2009

### kapv89

in MCRF of a body, the four accn has only spacial components. now since the body begins from rest in our frame, and since the four accn is of a constant direction and magnitude, we can take our x-axis to be along the direction of the four accn when MCRF of body is same as our frame. So now the problem is in 1+1 dimensions

Take our frame as O. in O the four accn of a body at a specific v and v(dot), as u gave, is
$$a^\mu=\left(\begin{array}{c}\gamma\dot{\gamma}c \\ \gamma\dot{\gamma}\mathbf{v}+\gamma^2\dot{\mathbf{v}}\end{array}\right)=\left(\begin{array}{c}\gamma\dot{\gamma}c \\ \gamma\dot{\gamma}\mathbf{v}+\gamma^2\mathbf{a}\end{array}\right)$$

In the MCRF of the body, the 4-accn has only a spatial component which should be of magnitude alpha(as given in the problem) . the MCRF moves at a speed v in x dir w.r.t. O, make a lorentz transformation of four accn from MCRF to O, and equate any of the component with the corresponding one in ur expression

Conversely, make a lorentz transformation of ur expression of four-accn(which is in O) to the MCRF, you will just have one component to compare then

8. Dec 16, 2009

### jdwood983

In a co-moving frame, $\gamma=1$ and $\dot{\gamma}=0$, so then you have $a^\mu=(0,\mathbf{a})=(0,a_x,a_y,a_z)$ while $v^\mu=(1,0,0,0)$--using $c=1$ (see page 51 of your text), so I'm not sure that you can make that argument.

9. Dec 16, 2009

### kapv89

i am not clear which argument are you talking about. can you please explain it

Last edited: Dec 16, 2009
10. Dec 16, 2009

### jdwood983

Specifically I was talking about looking at $a^\mu$ in the co-moving frame & comparing it to $v^\mu$ in the co-moving frame (they are orthogonal in this frame, so it is useless). But overall, I disagree with your choice of setting $\mathbf{v}=v\hat{\mathbf{x}}$ because the problem does not say that it is. I would suggest looking at $a^\mu$ and $v^\mu$ in the $\mathcal{O}$ frame.

11. Dec 16, 2009

### kapv89

i too would have preferred working with general coordinates but first i think i need to learn how to derive lorentz transformation for arbitrary velocity(all 3 components), can u recommend something?

but coming to how i worked on things, in the problem it is mentioned that body starts from rest, let the frame in which the body was at rest/from which it starts be our frame, i.e. O. now at t=0, when body is starting, the MCRF is same as O as the body is at rest wrt O. And in MCRF, 4-accn has only spatial components (which remain the same throughout the body's path as 4-accn is given const). now the direction in which this four accn at t=0 is a spatial direction. Take ur x-axis along that direction, and 4-accn has magnitude alpha and direction x. also given in the book is that this case is similar too that off a rocket with const. accn, and that 4 accn in mcrf is equivalent to galilean accn. for a very small time dt, the new velocity of object in O would be, by einstein's velocity addition formula, (alpha)dt (as initial was 0) 'the' x-direction. keep using velocity addition by moving to MCRF, looking at direction of four accn, taking initial velocity equal to the the velocity at which the MCRF was taken. Nowhere is a change in direction of movement obsvd. This is what i thought in concept when thinking whether or not i can take that 'x' direction. And also by comparing it with a rocket with const accn in x direction. I did'nt come across any equations which say its spatial direction would change. I think i will derive it this way too now

Also, I have never talked of comparing 4-accn and four velocity, i just said compare the components of the 4-accn given in the question and the expression you gave by applying proper lorentz transformation

12. Dec 16, 2009

### jdwood983

You are given that $\mathbf{a}$ is constant in direction as well as magnitude ($\mathbf{a}\cdot\mathbf{a}=\alpha^2\geq0$), you can't just assume that this direction is only in the $x$-direction. If there is an acceleration component in the $a_y$ (or $a_z$) direction, the motion clearly will not be just in the $x$-direction.

Again, $v^\mu=(1,0,0,0)$ in the MCRF, so you can't have any spatial components of velocity in the MCRF. The MCRF means that you are moving with the body such that the body's velocity, to your eyes, is zero. The problem is asking for the body's velocity in the earth-observer frame (that is, not in the MCRF). I realize that part (a) of the problem involves the MCRF, but this part of the problem does not.

So you want to just compare the co-moving frame acceleration to the observing frame acceleration? I don't believe this will work either because you are trying to compare the acceleration in two different reference frames.

13. Dec 16, 2009

### kapv89

for the first point, Even if the accn has ay and az components, we can take our x, or y or z axis to point in the direction in which accn points. just like rotation of x-y plane to desired direction changes nothing in z-axis, rotation of x-y-z space to suit my purpose won't change anything in t

for the second point, in the argument i gave was just trying to mathematically justify the fact that a rocket with a constant accn won't change its direction. and if starting from rest, in SR, will just move in the direction of its accn

the velocity addition thing works like this (i am having some problem with this method, have left it for later). when the speed of body is v in O, look in the MCRF. for a very small time dt in MCRF, let the body acclerate according to the accn vector (since we are still in MCRF, we can use the time coordinate instead of proper time). The speed of body now in O can be found out using einstein's velocity addition. its a bit complicated in solving cos i think something should be done with the dt, i have left it for later anyways

For the third point, I have read the theory and derivations till end of chapter 8(jumped the line :P) and it has been mentioned a lot of times in the book that vectors, one forms, tensors exist regardless of our coordinate frame, coordinates just describe them (think of a vector in a plane and rotate the coordinate axis, the vector is same, the components change as basis vectors change). By applying proper lorentz transformation, the four accn given in ques can be changed to that in frame O (by taking its value in MCRF from the stuff given in question). Now this value in O should be equal to the expression of 4 accn which u gave. this gives the answer

14. Dec 16, 2009

### jdwood983

Okay, this is fine. What I gathered from your previous posts that you wanted to look at just the $x$-direction of $a^\mu$ and that (my interpretation) is not valid.

My whole point is that you don't want to be looking at the co-moving frame because you have no velocity in the co-moving frame. This problem should be solved looking only at the $\mathcal{O}$ frame.

15. Dec 16, 2009

### kapv89

yeah i get that, but if u convert the expression for 4-accn in O to that in MCRF, your time component becomes 0 in that expression and all u are left with is the spacial part which can be put into an equation like this (gamma^3)(v(dot))= alpha (this is the same equation u get when u do it other way, i.e. convert given four accn from MCRF to O). Now since we have a relation for v, we can solve it

16. Dec 16, 2009

### jdwood983

In the co-moving frame, $a^\mu=(0,a_x,a_y,a_z)$ and $\gamma=1$ while in $\mathcal{O}$, $a^\mu=(\gamma\dot{\gamma}c,\,\gamma\dot{\gamma}\mathbf{v}+\gamma^2\dot{\mathbf{v}})$. So, again, I'm not sure you can make the previous argument. You can, however, use Equation (2.24) in conjunction with your equation from Post #3:

$$\dot{\gamma}=\frac{d\gamma}{dt}=\gamma^3\mathbf{v}\frac{d\mathbf{v}}{dt}$$

Separate the result for $d\mathbf{v}/dt$ and then integrate. This is how I would go about the problem.