Inertial observers measuring an object's acceleration...

[KipIngram]

Hi. I decided last night to work through the derivation of the Lorentz transform. I've "learned about it" in the past, but I thought it would do me good to pay full and rigorous attention to the math each step of the way. So I started off, and before long had the the transform that relates the coordinates of an event (x, ct) for observer S and (x', ct') for observer M when S sees M moving at constant velocity v. So far so good. I had it written in matrix form, so I inverted it and confirmed that the transform and its inverse gave me the identity. Score one for the home team.

 

Then I looked at the situation where S observes an object moving with constant velocity U, and worked out that M would also observe the object moving with a constant velocity U' and found the relationship between U' and U. Score again.

 

Then I tinkered with S observing an object that was at rest at his x=0 for t<0, but at t=0 initiated uniform acceleration A. That ran me into trouble. I tried noting that at t=2 S would see the object at x=2A, giving me an event with coordinates (2A, 2c), and then transformed that into M's frame. I reasoned that for t<=0 M would have seen the object moving at velocity -v, an initial velocity, and then would see uniform acceleration with some value A'. Using the above mentioned transformed event, I had a location and a time that M would associate with the object, an tried to plug that into M's dynamic equation for uniform acceleration, x' = A't'^2/2 - vt' so I could solve for A' in terms of A and usual factors. It just went nowhere.

 

I may have an error in my algebra that I haven't caught yet. But it also occurred to me that I wasn't sure whether inertial observers will agree as to whether an object is undergoing uniform acceleration.

 

So that's my question. Do they, or do they not? I'd rather not chase an algebra error if the problem is conceptual.

 

Somewhere upstairs I have a book that went through stuff like this, but I was hoping to work as much of this out as I could myself without going to a reference; I'll remember it better if I do.

 

Thanks!

 

Edit shortly later:

 

I think I convinced myself that they do not. S sees velocities 0, A, 2A, 3A, etc. at one second intervals. Those intervals are not one second intervals for M, but they are fixed size intervals. My expression for velocity transformation was U' = (U-v) / (1 - uV/c^2). The numerator of that changes by fixed amounts each "tick," but the denominator is changing too, so the steps in U' are not the same. So it looks like M will observe non-uniform acceleration.

Edited May 13, 2017 by KipIngram

[koti]

Wouldn't they both agree that the object is undergoing uniform acceleration just measure different rates of it?

[KipIngram]

My original assumption was exactly that - that the two observers would each see constant acceleration but different values.

 

I'm sort of tying my self in knots and going back and forth. My original plan was to identify the events that corresponded to the test object; if it starts with uniform acceleration A at t=0, then those would all have the (x, ct) form (At^2/2, ct). Those are good events that I can use the Lorentz transform on to find out what (x', ct') events you'd associate with them and thus have the x' vs. t' curve that you would measure. Bearing in mind that you see the test object with an initial velocity -v, then if you also see uniform acceleration we ought to be able to plug some of those values into x' = A't'^2/2 - vt' and arrive at a good value for A'. If you don't see uniform acceleration, then that wouldn't work.

 

I wound up with non-working algebra when I tried that, but like I said earlier I may have just dropped the ball somewhere. I was trying to do it in a console editor session, so it wasn't the most friendly "visual" environment for such things. And my method presumed from the start that a single constant value for A' would result, so if that's not right it was bound to fall apart. But at any rate, I was finding it problematic.

 

Then when I wrote the follow-on note I was focusing on velocities. I noted that I saw velocity advance by A each second, but that when I applied the velocity transformation rule I'd derived early (and confirmed online), I got non-uniform steps for the transformed velocity. That combined with the fact that our clocks do run at a fixed and unchanging ratio with respect to one another made me decide that you would not see a uniformly increasing velocity.

 

But now I'm questioning that; I feel like maybe taking that simpler approach glosses over something. I'm not fully decided either way yet. I feel like the first approach I outlined above is the most solid and sure one, so I may make another pass at it without initially assuming your acceleration will turn out constant. It's a valid approach as long as we keep it general enough to handle either outcome.

Ok, I think I showed myself that M's measurement will not yield uniform acceleration. Rather than slug through the algebra, I chose a specific numerical case. I set c=1, took v=beta=0.5, and took A to be 0.001c numerically. I solved for three different transformed event points and substituted those into what should be M's equation of motion solved for A'. I got different values in all three cases. The difference wasn't large, but I carefully carried all available digits of accuracy on my calculator , and I did it for three points (2, 4, and 6 seconds) and got a monotonically increasing value for A'.

 

I don't know that I consider this final - I've done a good bit of numerical analysis in my time and while I'm pretty good at it I also know it can burn you if you aren't really, REALLY careful. My expressions were precise up until pretty near the end, so there wasn't long for round off error to accumulate.

 

I'm hoping someone more knowledgeable will weigh in.

In case anyone is interested, here is the analysis I just did. I use k for (1-v^2/c^2)^0.5 and b for v/c.

 

4. Acceleration. Let S observe a test object that is resting at (x,t) = (0, 0) and then

initiates uniform acceleration A. S will associate these events with the object:

E2 = (2A, 2c)

E4 = (8A, 4c)

E6 = (18A, 6c)

All of these events can be translated into M's frame.

E2' = (2Ak-2bkc, 2kc-2Abk)

E4' = (8Ak-4bkc, 4kc-8Abk)

E6' = (18Ak-6bkc, 6kc-18Abk)

Assume A = 0.001c. Then

E2' = (0.002ck-2ckb, 2ck-0.002ckb)

E4' = (0.008ck-4ckb, 4ck-0.008ckb)

E6' = (0.018ck-6ckb, 6ck-0.018ckb)

This means that the (x', t') points at which M will observe the object are

(0.002ck-2ckb, 2k-0.002kb)

(0.008ck-4ckb, 4k-0.008kb)

(0.018ck-6ckb, 6k-0.018kb)

Let us further assume that c=1 and b=v=0.5. Then

(0.002k-1k, 2k-0.001k) = (-0.998k, 1.999k)

(0.008k-2k, 4k-0.004k) = (-1.992k, 3.996k)

(0.018k-3k, 6k-0.009k) = (-2.982k, 5.991k)

--------------------------- I think it's precise up to here, and the rest was carried a full calculator precision.

Finally, for b=0.5 we have k = 2/(3^0.5) = 1.1547, so M's observation points are

(-1.1524, 2.3082)

(-2.3002, 4.6142)

(-3.4433, 6.9178)

Now note that M observes the object to have an initial velocity of -v, so if he is

to observe uniform acceleration it would fit this equation of motion:

x' = A't'^2/2 - vt'

For any single data point (except t'=0) we can solve for A':

A' = 2(x'+vt')/(t'^2)

Recalling that v=0.5 and subtituting the values above for x' and t', we get

A'(E2) = 6.501691e-4

A'(E4) = 6.508200e-4

A'(E6) = 6.514720e-4

When I was toying with the direct transformation of observed velocities I felt it implied that the acceleration seen by M would increase (at least for this case).

Edited May 14, 2017 by KipIngram

[Mordred]

There is several steps to handle uniform acceleration. This involves rapidity. This paper details the required steps.

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.physik.uni-leipzig.de/~schiller/ed10/Uniform%2520relativistic%2520acceleration.pdf&ved=0ahUKEwj4__7oyO7TAhVE5mMKHb0HAT8QFggnMAM&usg=AFQjCNE8A2T74ijQXQJHYZLeaKuSLK5roQ&sig2=Cj-Fu2-U5nxBCYf5dDQjaw

In particular the section on hyperbolic spacetime paths.

equations 2.8 to 2.10 are of particular note.

As you noticed the basic Minkowskii matrixes and Lorentz boost involve constant velocity. When you undergo acceleretion you need to calculate the rapidity under 4 momentum 4 velocity. The groups handle this under rotations. (in essence you need to use instantaneous velicities) These examples will help

I'll dig around for better examples the ones I have are in my textbooks.

Equation 18 here is your uniform acceleration (without change in direction)
transformations and how they derive it.

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://aether.lbl.gov/www/classes/p139/homework/eight.pdf&ved=0ahUKEwj4__7oyO7TAhVE5mMKHb0HAT8QFggqMAQ&usg=AFQjCNEp_TbZhvDInabh8on8EEslpMnQOg&sig2=_yNYf6cM_Vvva4V_bbG5_A

Here is a workup I've posted in the past as another example this is the twin paradox turnaround. (change of direction =acceleration change but not constant). The above articles detail uniform acceleration.

 

The transforms with rapidity are as follows.

[math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math]

The formula you posted are Lorentz but only under constant velocity. The above matrix provide the changes due to rapidity. Some articles may refer to it as your hyperbolic rotation.

first define the four velocity. [latex]u^\mu[/latex]
[latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex]

this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex]
the four velocity has constant length.

[latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex]

the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex]
[latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex]

so the acceleration and velocity four vectors are

[latex]c \frac{dt}{d\tau}=u^0[/latex]

[latex]\frac{dx^1}{d\tau}=u^1[/latex]

[latex]\frac{du^0}{d\tau}=a^0[/latex]

[latex]\frac{du^1}{d\tau}=a^1[/latex]

both vectors has vanishing 2 and 3 components.

using the equations above
[latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex]

in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here?

the last equation defines constant acceleration g. with solutions

[latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex]

from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex]
hence
[latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex]
similarly
[latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex]

so the solution to the last equation is
[latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex]
hence
[latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex]

with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex]

so
[latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]
hence
[latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]
and finally
[latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex],
[latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex]

the space and time coordinates then fall onto the Hyperbola during rotation
[latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex]

Edited May 14, 2017 by Mordred

[KipIngram]

Thanks Mordred; I'm reading. So, where did the approach I was taking fall down? I think I feel confident about taking a set of specified events in the S frame and moving them to the M frame - that's exactly what the Lorentz transform is supposed to do (connect those (x, ct) <--> (x', ct') pairs).

 

That leaves me with the very presumption that the equation of motion x = At^2/2 + v(0)*t holds; is that itself something that's not strictly correct?