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Eroica
2004-Mar-31, 01:56 PM
On another thread (http://www.badastronomy.com/phpBB/viewtopic.php?p=228155#228155) we were looking at the general equations for a ship that is travelling through space with constant jerk (jerk = rate of change of acceleration), and someone wondered what they would be like if relativistic considerations were included in the calculations. Here's how far I got.

There are two frames of reference: an inertial one that is at rest, and the ship itself. Inertial time and displacement are t and s respectively. The corresponding dilated time and contracted displacement as measured by the crew of the ship are t' and s'. At time t = 0, let us assume that s, v and a = 0. The ship's jerk = j, a constant.

The equations we derived are (all parameters relative to the inertial frame):

a = jt .... Equation 1

v = ½jt² .... Equation 2

s = (1/6)jt³ .... Equation 3

In order to calculate t' and s', I used the gamma factor = √[1-(v²/c²)]. If you plot the gamma factor against undilated time, the area under the graph is the corresponding dilated time. In other words:

t' = ∫γdt

Similarly:

s' = ∫γds

Let's take time first. Starting with Equation 2:

v = ½jt²

=> v² = ¼j²t^4

=> v²/c² = (j²t^4)/4c²

=> 1-(v²/c²) = (4c²-j²t^4)/(4c²)

=> γ = (1/2c)√[4c²-j²t^4]

=> t' = (1/2c)∫√[4c²-j²t^4]dt

Trouble is, this isn't a standard integral and I haven't been able to solve it yet. Any ideas?

As for s', it seems to be even more complex:

s = (1/6)jt³

=> ds = ½jt²dt

=> s' = (¼c)∫√[4c²-j²t^4]t²dt

with the same problem as before. ](*,)

[Edit: The last equation has an error. it should read:

s' = (j/4c)∫√[4c²-j²t^4]t²dt

Thank you, Gsquare, for highlighting this error.]
[Second Edit. I changed the notation to clarify things.]

nitzer219
2004-Mar-31, 02:18 PM
Eroica -

Question:

Let's assume that neither the transmission (video feed to ground control (gc)), nor the ship can actually reach c, since as far as we know now, they can't.

Is this then, a correct assumption?

1. The shuttle, fractionally, exceeds the speed of the transmission.
2. The transmission would not equal 'c', as it's not in a complete vacuum (there is matter in space, so c is not going to be attained totally).
3. If the ship exceeds the speed of the transmission, would not the transmission fail to ever reach earth again, at the point the shuttle surpasses the speed of the transmission?

I'm still waiting on a formal reply from the guru on this...I have to believe this is solvable. Thanks for seeking out feedback for the answer.

Although I thought this nightmare up, I've never had the mathematical experience, practice, or knowledge to prove it...so I'm in my own kind of hell with this one...and what the heck..I figured...why not share my hell with others. :evil:

informant
2004-Apr-01, 01:12 PM
Trouble is, this isn't a standard integral and I haven't been able to solve it yet. Any ideas?
Those are the integrals of rational functions of t and a+b*t^4 (a=4*c^2, b=-j^2). You should be able to find the appropriate substitution in a calculus text, but, if my quick glance is right, neither integral can be expressed in terms of elementary functions.

George
2004-Apr-02, 07:40 PM
Those are the integrals of rational functions of t and a+b*t^4 (a=4*c^2, b=-j^2). You should be able to find the appropriate substitution in a calculus text, but, if my quick glance is right, neither integral can be expressed in terms of elementary functions.

I'll bet I'm wrong but figured I'd ask to learn something here.

Aren't a &amp; b restricted to being constants? If so, a is fine but b is not a constant.

informant
2004-Apr-02, 07:44 PM
Look at the OP. Eroica started by assuming that j was constant.

George
2004-Apr-02, 10:44 PM
Look at the OP. Eroica started by assuming that j was constant.

:roll: Yep. That explains pretty well. Shouldn't be hard at all now.

skrap1r0n
2004-Apr-02, 11:01 PM
darn I clicked on this thread hoping you were going to be talking about Al Franken... :roll:

countrywideoptionone
2004-Apr-03, 12:20 AM
darn I clicked on this thread hoping you were going to be talking about Al Franken... :roll:

Hahaha. Are you saying that Al is a third derivative of a position?

skrap1r0n
2004-Apr-03, 12:50 AM
darn I clicked on this thread hoping you were going to be talking about Al Franken... :roll:

Hahaha. Are you saying that Al is a third derivative of a position?

I would be lying if I said I had a clue > 0 as to what this thread is about. I meant the title. I'll assume a relativisic jerk is a third derivitive of a position? :D

informant
2004-Apr-03, 01:36 PM
Eroica, you should be able to expand the integrand in a Taylor series. That could be enough for a thought experiment.

Eroica
2004-Apr-03, 04:25 PM
Let's call that Plan B! :D

Seriously, though, thanks for the tip.

Gsquare
2004-Apr-04, 02:42 AM
=> t' = (½c)∫√[4c²-j²t^4]dt

=> s' = (¼c)∫√[4c²-j²t^4]t²dt

Well, Eroica; good try, but solving these eqns. won't get you very far. [-X

The first indication that these equations are incorrect is the fact that s' differs from t' by a factor of only 1/2 t^2. Does that give you a clue?

The last I checked, 'time' is still measured in seconds and s (distance) is in meters; (unless they've changed that too since I was in college!) :wink: ....so unless there is a new transformation eqn. showing how to turn seconds into meters simply by multiplying by t^2 then something is certainly amiss; (I'm sure one of my old college professors has a way to do it, but then again it probably requires taking drugs). He, he, just kiddin'. Your t' equation is probably not in seconds anyway; both are probably in incorrect units.

If you dont mind me saying so.The root of the problem arises from 1). the methodology you are using and 2). the inapplicability of trying to use an SR derived equation in a non-inertial frame.

The equation you are trying to use for time dilation is an SR derived eqn. strictly for comparing inertial frames. For NON-inertial (accelerating) frames the gravitational time dilation formula is usually used (which, if Im not mistaken, is a factor of:
(1 + 2U/c^2), where U = -GM/r. (Or use the equivalent accelerating eqn.)

Howeeeeever, since your acceleration is NOT constant, then you will most likely need to derive a new jerkie time dilation formula which will probably require full blown GR since it will be equivalent to a changing gravitational field (and it wont be as simple as integrating the gravitational time dilation eqn. :P).

No I cant, I wont, I shall not help you! [-( It seems to me Ive seen someone do this a few years ago; however, due to time dilation it has long ago escaped me.

Nevertheless, you seem very capable of handling it. Hey, you may even get gravitational waves! This is reasonable since jerk is equivalent to a CHANGING gravitational field. All you need to do is arrange the mass distribution to provide for quadrupolar emission. \:D/ :D Now theres an idea you can work on. =D> :D :P

G^2 :wink:

Eroica
2004-Apr-04, 09:23 AM
Thanks for the input, Gsquare.

If you dont mind me saying so.The root of the problem arises from 1). the methodology you are using and 2). the inapplicability of trying to use an SR derived equation in a non-inertial frame.

The equation you are trying to use for time dilation is an SR derived eqn. strictly for comparing inertial frames. For NON-inertial (accelerating) frames the gravitational time dilation formula is usually used (which, if Im not mistaken, is a factor of:
(1 + 2U/c^2), where U = -GM/r. (Or use the equivalent accelerating eqn.)
You are quite wrong when you say that SR cannot handle acceleration:

Can Special Relativity Handle Accelerations? (http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html)

I belive my methodology is correct and valid. There are no gravitational fields involved in the problem, therefore GR is not relevant. What SR cannot handle is gravity, not acceleration or jerk. This is probably the commonest misunderstanding in relativity. (I was guilty of it myself until this board set me straight some months ago.)

The first indication that these equations are incorrect is the fact that s' differs from t' by a factor of only 1/2 t^2 ... The last I checked, 'time' is still measured in seconds and s (distance) is in meters...
Thanks for highlighting an error. I have rechecked my equations, and I did indeed make one crucial mistake in the equation for s' in the OP. It should read:

s' = (j/4c)∫√[4c²-j²t^4]t²dt

I left out the first j. Obviously jt² has the dimensions m/s. Multiply a time by m/s and you get a displacement. End of problem.

I stand over both (corrected) equations and my methodology.

[Edited for clarity]

shash
2004-Apr-04, 02:47 PM
I only did very little, but how about Laplace transforms? It should be pretty easy...

I'll try that myself.

shash
2004-Apr-04, 03:43 PM
OK, I tried t':

What I got was

t' = (½c) ??[4c²-j²t^4]dt

=> t' = (½c) ?j?(4c²/j²-t^4)dt (multiplying and dividing by j² in the root and taking j out)

=> dt'/dt = j * (½c) ?(4c²/j²-t^4) (Taking differential of the whole thing wrt t)

=> s(T') = j/2c * L (?(4c²/j²-t^4)) (The Laplace transform, t'(0) = 0, t(0) = 0)

=> s(T') = j/2c * exp{-s(4c²/j²)} * L( ?t^4) (by 1st shifting theorem of Laplace Transforms)

=> s(T') = j/2c * exp{-4c²s/j²} * L(t²)

=> s(T') = j/2c * exp{-4c²s/j²} *1/s^3

=> T' = j/2c * exp{-4c²s/j²} *1/s^4 (dividing both sides by s)

=> t' = j/2c * L'{exp{-4c²s/j²} *1/s^4} (Taking inverse Laplace)

=> t' = j/2c * L'{1/s^4} [t -> (t + 4c²/j²)] (Corollary of 1st shifting theorem)

=> t' = j/2c * t^3 [t -> (t + 4c²/j²)]

=> t' = j/2c * (t + 4c²/j²)^3

I don't know about the dimensional correctness, since I seem to be getting a t^3 for a t term.

But looking at your original equation, that seems to be the case there too... So, where's that heading?

EDIT: Somehow the integral and sqrt signs got messed up :( !

In the first step, the first ? is integral, and the second is sqrt. In the second and third, the ? is sqrt.

Eroica
2004-Apr-04, 04:34 PM
Thanks for the assistance, shash, and welcome to the board! =D>

In the original equation, 4c² and -j²t^4 have the same dimensions, viz m²/s². But in your final equation, you have the term (t + 4c²/j²)³. These two terms have different dimensions (s and s^4), so they can't be added, can they?

Unfortunately I know nothing about Laplace Transforms, so I don't know whether you have gone wrong somewhere or whether the original data I supplied were flawed.

(j, jerk, has the dimensions m/s³)

shash
2004-Apr-04, 05:31 PM
#-o I see what I did wrong!

The first shifting theorem in step 4 is not applicable if t has a power separate of the other term!

JohnOwens
2004-Apr-05, 05:45 AM
The last I checked, 'time' is still measured in seconds and s (distance) is in meters; (unless they've changed that too since I was in college!) :wink: ....so unless there is a new transformation eqn. showing how to turn seconds into meters simply by multiplying by t^2 then something is certainly amiss; (I'm sure one of my old college professors has a way to do it, but then again it probably requires taking drugs). He, he, just kiddin'. Your t' equation is probably not in seconds anyway; both are probably in incorrect units.
While you were right (I think?) in this case, that doesn't hold in relativity equations in general. There are certainly equations where time and distance are used equivalently, with c as the ratio between the two.

milli360
2004-Apr-05, 09:02 AM
The last I checked, 'time' is still measured in seconds and s (distance) is in meters; (unless they've changed that too since I was in college!) :wink: ....so unless there is a new transformation eqn. showing how to turn seconds into meters simply by multiplying by t^2 then something is certainly amiss; (I'm sure one of my old college professors has a way to do it, but then again it probably requires taking drugs).
No drugs necessary. :)

Did you know the mass of the Sun (http://www.badastronomy.com/phpBB/viewtopic.php?p=51760#51760) was about a kilometer and a half?

Gsquare
2004-Apr-05, 03:28 PM
[quote=Gsquare]

No drugs necessary. :)

Did you know the mass of the Sun (http://www.badastronomy.com/phpBB/viewtopic.php?p=51760#51760) was about a kilometer and a half?

Yea, you just made my point, see what geometrized units makes one sound like? :lol: :D
No drugs neccessary is right.... just use geometrized units. :wink: :D

Luckily I was out of college before all this nonsense started. I like seeing &amp; knowing &amp; remembering what I'm dealing with.
c is still = 1/sq.rt.(eu), NOT = 1. :roll:

G^2

milli360
2004-Apr-05, 04:05 PM
Yea, you just made my point, see what geometrized units makes one sound like?

So what point did I make, again?

Luckily I was out of college before all this nonsense started.
I find it in MTW's bible of relativity, Gravitation, which was published in 1971 I think. It goes back further than that. You seem much younger.

George
2004-Apr-05, 04:20 PM
...The equation you are trying to use for time dilation is an SR derived eqn. strictly for comparing inertial frames...
You are quite wrong when you say that SR cannot handle acceleration:

Can Special Relativity Handle Accelerations? (http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html)

I belive my methodology is correct and valid. There are no gravitational fields involved in the problem, therefore GR is not relevant. What SR cannot handle is gravity, not acceleration or jerk. This is probably the commonest misunderstanding in relativity. (I was guilty of it myself until this board set me straight some months ago.)...

I certainly am not qualified to argue but I am wondering how valid this is. The link had a couple of statements that trip me up (a common experience for me :roll: ).

...Special relativity gives a completely self-consistent description of the mechanics of accelerating bodies neglecting gravitation, just as Newtonian mechanics did.

The formalism is very similar to that of many general relativity problems but it is still special relativity so long as the space-time is constrained to be flat and Minkowskian...

Doesn't "equivalence" disallow such ideas?

Won't flat space-time limit reality?

shash
2004-Apr-06, 04:50 PM
Only if you're dealing with immense gravities.

If our hypothetical spacecraft were to encounter a star, flat space-time would become a problem, and it'd be better to use curved space. Otherwise, space-time is flat enough to use SR anyway, no matter how many "layers" of velocity change you add (acceleration, jerk, push, shove or whatever name you care to give them)...

George
2004-Apr-06, 06:36 PM
Only if you're dealing with immense gravities.

If our hypothetical spacecraft were to encounter a star, flat space-time would become a problem, and it'd be better to use curved space. Otherwise, space-time is flat enough to use SR anyway, no matter how many "layers" of velocity change you add (acceleration, jerk, push, shove or whatever name you care to give them)...

Thanks. I was assuming a "Big Jerk" would imply a "big problem". :wink:

countrywideoptionone
2004-Apr-07, 10:53 PM
darn I clicked on this thread hoping you were going to be talking about Al Franken... :roll:

Hahaha. Are you saying that Al is a third derivative of a position?

I would be lying if I said I had a clue > 0 as to what this thread is about. I meant the title. I'll assume a relativisic jerk is a third derivitive of a position? :D
It's a joke that floats around calculus classes. Take any position (x,y), on a curve. If you interpret the graph as x - time, y - distance. Average speed is d/t, or in this case (change in y)/(change in x), which is also how you find the slope of a line. What if it's not a line you're looking at, but a curve? How do you find the slope now? If the curve is "smooth," you can zoom in so far that it appears to be a straight line. Now you can find the slope, but the slope can be different at each point on the curve, so a more effiecent way is to find a related equation that represents the slope of the original equation. This new equation is called a derivative (it is "derived from").

Here's a simple example of a derivative. f(x)=x^2 is your plain old parabola. What is it's slope? Well, it depends on what point you want to look at. You could continue to zoom in on one point until your field of view resembled a straight line, and then pick to points "very close" to your point, and use the old slope formula, or via calculus, you can derive the slope equation f(x)'=2x. So to find the slope of f(x), at a point, put the x value into f(x)'. Ex: what is the slope at (4,16)? f(4)'= 2(4)=8.

...I digress. So if f(x) is your curve representing the distance of something to (0,0), f(x)' (the derivative) what represents the rate of change in distance (position) which is the speed.

But what if the speed is not constant? Then it changes! So, we now know that to find the rate of change of a curve, we can find the derivative. So, we take the [derivative of f(x)'], or f(x)'', also called the second derivative, or in this case, the change in speed, which we can call acceleration.

But what if the acceleration is not constant? Then it changes! So, we take the derivative of f(x)'', called the third derivative, or f(x)'''. But what is this really? Have you ever been driving in your car, and you're lightly pressing the brake, so that your acceleration in fairly constant? Now, during this slowing down process, you push on the brake much harder. Your body feels like it's being thrown forward doesn't it? You could call this a jerk. So you could say that the third derivative of a position is a jerk. Alternatively, if someone is being "a jerk" you could tell them, "Hey, stop being a third derivative of a position!"