# A question about v appearing in the transformation equations in SR

1. Mar 23, 2014

### ftr

2. Mar 23, 2014

### Staff: Mentor

The v is the speed of the origin of one frame in the coordinates of the other frame.

3. Mar 23, 2014

### ftr

I understand that part, but mathematically we can write v as differentials with appropriate symbols, what are they?

4. Mar 23, 2014

### Staff: Mentor

In any frame, the velocity of any point is dx/dt, where x is the x coordinate of the point in that frame as a function of t and t is the time coordinate in that frame.

If I have two frames (primed and unprimed) and in the unprimed frame the origin of the primed frame is moving in the positive x direction with speed v, I'll write the coordinates of the origin of the primed frame as (x=vt, t=t) and dx/dt is v.

5. Mar 24, 2014

### georgir

I'm guessing OP is wondering for the cases with more than one space dimension...

6. Mar 24, 2014

### Staff: Mentor

The position of the origin of one frame written in the coordinates of the other frame is $x=v t$. So v is the standard $v=dx/dt$.

Last edited: Mar 24, 2014
7. Mar 24, 2014

### Staff: Mentor

Could be... If so, he'll come back and tell us and then we can provide the three-spatial-dimensional answer and/or the coordinate transform that reduces the three-dimensional problem to the one-dimensional problem for which DaleSpam and I supplied the answer.

As an aside... This a is question of classical Galilean relativity.

8. Mar 24, 2014

### ftr

Then the problem I have is when I substitute delta x and delta t in the transformation using the expressions below (from wiki), then you get v appearing again as an infinite recursive relation. Not sure how that works.

we get
\begin{array}{ll}
\Delta x' = \gamma \ (\Delta x - v \,\Delta t) \ , & \Delta x = \gamma \ (\Delta x' + v \,\Delta t') \ , \\
\Delta t' = \gamma \ \left(\Delta t - \dfrac{v \,\Delta x}{c^{2}} \right) \ , & \Delta t = \gamma \ \left(\Delta t' + \dfrac{v \,\Delta x'}{c^{2}} \right) \ . \\
\end{array}

9. Mar 24, 2014

### Staff: Mentor

Why is there any recursion? The v for the forward transform is the same as -v for the backwards transform, as you showed. If the origin of the primed frame is moving at v in the unprimed frame then the origin of the unprimed frame is moving at -v in the primed frame. No recursion is needed.

Last edited: Mar 24, 2014
10. Mar 24, 2014

### ftr

eq 1

\begin{align}

x' &= \gamma \ (x - v t) \\

\end{align}

eq 2

\begin{array}{ll}
\Delta x = \gamma \ (\Delta x' + v \,\Delta t') \ , \\
& \Delta t = \gamma \ \left(\Delta t' + \dfrac{v \,\Delta x'}{c^{2}} \right) \ . \\
\end{array}

I am substituting the ratio of delta x/delta t in equation 2 into v of equation 1

It is clear that you will see v again, and substituting for that you get v's again and so on.

11. Mar 24, 2014

### Staff: Mentor

Now I'm not sure what you're trying to show here... are you trying to derive the relativistic velocity addition rule that relates the value of the velocity dx'/dt' of an object in the primed frame to the velocity dx/dt of the same object in the unprimed frame?

Or are you just trying to solve
$$x' = \gamma(x-vt) \\ t' = \gamma(t-vx)$$

for $x$ and $t$ to get the expected
$$x = \gamma(x'+vt') \\ t = \gamma(t'+vx')$$
that tells us that the velocity of the primed frame relative to the unprimed frame is the negative of the velocity of the unprimed frame relative to the primed frame?

12. Mar 24, 2014

### Staff: Mentor

Why would you do that?

The origin of the primed frame in its own coordinates is obviously $x'=0$. There is no recursion or transformation involved, it is a definition. The origin of the primed frame in the unprimed coordinates is $x=vt$ which can be obtained from the transform.

The origin of the unprimed frame in its own coordinates is obviously $x=0$. There is no recursion or transformation involved, it is a definition. The origin of the unprimed frame in the primed coordinates is $x'=-vt'$ which can be obtained from the transform.

Why are you doing any recursion?

Last edited: Mar 24, 2014
13. Mar 24, 2014

### robphy

It might help to label events.

\begin{align}

x_A' &= \gamma \ (x_A - v t_A) \\
x_B' &= \gamma \ (x_B - v t_B) \\

\end{align}

\begin{array}{ll}
\Delta x_{\mbox{A to B}} = \gamma \ (\Delta x'_{\mbox{A to B}} + v \,\Delta t'_{\mbox{A to B}}) \ , \\
& \Delta t_{\mbox{A to B}} = \gamma \ \left(\Delta t'_{\mbox{A to B}} + \dfrac{v \,\Delta x'_{\mbox{A to B}}}{c^{2}} \right) \ . \\
\end{array}

Note, in general, these [point]events A and B need not have anything to do with v (the [slope]velocity of the worldline of the other observer in the diagram of the first observer).

14. Mar 24, 2014

### ftr

I just did the manipulation to see what results I will get. Obviuosly it was something not clear in my understanding. Of course I do understand the standard transformation.

I was not trying to do recursions, only the the manipulation seemed to suggest it.

Thanks. I think this is a bit clearer. However, it is still unclear somewhat. I have to think it over.

15. Mar 24, 2014

### Fredrik

Staff Emeritus
A Lorentz transformation describes a change of coordinates from a coordinate system S to another coordinate system S'. Coordinate systems assign coordinates to events. If some event p is assigned coordinates (t,x) by S, and (t',x') by S', then (in units such that c=1)
\begin{align}
\begin{pmatrix}t'\\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t\\ x\end{pmatrix}
\end{align}The v here has nothing to do with the event p. It is the velocity of S' in S. v is the "x/t" of an event on the S' axis, not the "x/t" of the event p.

To be even more precise, for all events p, let $(t_p,x_p)$ denote the coordinate pair that S associates with p, and let $(t'_p,x'_p)$ denote the coordinate pair that S' associates with p. Let q be an arbitrary event such that $x'_q=0$ and $t'_q\neq 0$. (This is an event on the S' axis). Then for all events p,
\begin{align}
\begin{pmatrix}t'_p\\ x'_p\end{pmatrix}=\frac{1}{\sqrt{1-\left(\frac{x_q}{t_q}\right)^2}}\begin{pmatrix}1 & -x_q/t_q\\ -x_q/t_q & 1\end{pmatrix}\begin{pmatrix}t_p\\ x_p\end{pmatrix}
\end{align}

Last edited: Mar 24, 2014
16. Mar 25, 2014

### dauto

You have to understand there are three objects in these transformations. There is an observer O, an observer O', and a Particle P. x and t are the coordinates of the particle P as seen from O. x' and t' are the coordinates of the particle P as seen from O'. v is the speed of O' as seen from O and has nothing to do with x or t. It makes no sense to differentiate x with respect to t and expect to get v. They are different things. v is the speed of O' and x and t are coordinates of P.

17. Mar 25, 2014

### Staff: Mentor

I'm getting the feeling you are looking for a precise mathematical expression for v. This can be done in terms of a partial derivative:
$$v=\left(\frac{∂x}{∂t}\right)_{x'}$$
The physical interpretation of this is that observers in the S frame of reference focus their attention on a specific location x' in the S' frame of reference and record its location (as measured in S) as a function of t (as reckoned in S). They then take the derivative of x with respect to t (for constant x'). This is what they regard as the velocity of the S' frame with respect to the S frame.
Another relationship for the same velocity involving a partial derivative as reckoned by observers in the S' frame is:
$$v=-\left(\frac{∂x'}{∂t'}\right)_{x}$$

The 3D version is:

$$v_x=\left(\frac{∂x}{∂t}\right)_{x',y',z'}$$

etc.

Hope that this is what you were looking for.

Chet

Last edited: Mar 25, 2014
18. Mar 25, 2014

### ftr

Thank you all for your help. I need to understand before I reply.

@Chestermiller

Yes indeed, you are very close to understanding my problem. Can you show that the two expressions you gave are equivalent, if it is not too much of a burden. Thanks in advance.

19. Mar 25, 2014

### Staff: Mentor

Sure. No problem.

x'=γ(x-vt)
So take the partial derivative of x with respect to t at constant x', and you get v.

x = γ(x'+vt')

So, take the partial derivative of x' with respect to t' at constant x, and you get -v.

Suppose you have a team of observers strung out along a railroad track, and you want to get the velocity of the train. So you focus on a specific guy on the train sitting at one of the train windows (x' = constant), and your team members write down the time at which this guy passes each of their locations (along with their coordinates along the track). Then, they get together later and plot a graph of their locations as a function of the time that the guy passed them. The slope of this line is the velocity of the train.

Now, there is a team of observers on the train strung out along the train, and they want to get the velocity of the train. So they focus on a specific guy on the ground (x = constant), and they write down the time that each of them passes this guy (along with each of their coordinates along the train). Then, they get together later and plot a graph of their locations as a function of the time that they each passed the guy. The slope of this line will be minus the velocity of the train (since the team members closer to the front of the train pass the guy first, and the team members further back pass him later).

Chet

Last edited: Mar 25, 2014
20. Mar 25, 2014

### Fredrik

Staff Emeritus
It seems that one of the things you want to know is what function v is a derivative of. I'll try to answer that.

The S and S' coordinate systems are the coordinates used by two observers O and O' respectively. No matter what coordinate system you're using, you would consider the world line of O' "the S' axis".

Let $x:\mathbb R\to\mathbb R$ be the function defined by saying that that for all real numbers t, the (time,position) coordinate pair in S, of the event on the world line of O' with time coordinate t, is (t,x(t)). By assumption, the velocity of O' relative to S is constant. This means that x''(t)=0 for all t. Also by assumption, O and O' were at the same place when both their clocks displayed zero. This means that one of the events on the S' axis has coordinates (0,0)=(0,x(0)). So we know that x(0)=0.

These conditions imply that there's exactly one real number r such that x(t)=rt for all t. Let's denote it by v. We have x(t)=vt for all t. Since x(t) denotes the position of O' at time t, x'(t) is the velocity of O' at time t. Since x'(t)=v for all t, we can say that at any event on the world line of O', the coordinate velocity of O' in S at that event is v.

21. Mar 28, 2014

### ftr

With proper substitution(using the same equations as in wiki) for v the equations are consistent.

22. Mar 28, 2014

### Fredrik

Staff Emeritus
Yes, I think what you have just observed is that if $\Lambda$ is a Lorentz transformation, we have
$$\begin{pmatrix}x'\\ t'\end{pmatrix} =\Lambda\begin{pmatrix}x\\ t\end{pmatrix}= \Lambda\Lambda^{-1}\begin{pmatrix} x'\\ t'\end{pmatrix} =I\begin{pmatrix} x'\\ t'\end{pmatrix}=\begin{pmatrix} x'\\ t'\end{pmatrix}.$$ It's slightly harder to see that the third equality holds when we use the explicit formulas for $\Lambda$ and $\Lambda^{-1}$ (this appears to be what you're doing in a different notation),
$$\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},\qquad \Lambda^{-1}=\frac{1}{\sqrt{1-(-v)^2}}\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}$$ but if you just multiply these two together, you will get the identity matrix.