# Susskind Lecture - accelerate reference frames

1. Sep 12, 2011

### csmcmillion

in

at 0:47:00 Susskind begins discussing accelerated reference frames and notes that they relate to hyperbolas rather than parabolas. I understand the concept and need for the proper acceleration to be asymptotic at C. Susskind seems to infer that an observer in the accelerated frame will experience constant acceleration forever ("observer feels the same amount of acceleration at all times"). How can this be so? It seems that the acceleration would fall off with time as the relativistic mass grows ever more rapidly, even for observers in the proper frame...

- Confounded

Last edited by a moderator: Sep 25, 2014
2. Sep 12, 2011

### TrickyDicky

He is referring to SR in that particular example.

Last edited by a moderator: Sep 25, 2014
3. Sep 12, 2011

### Bill_K

csmcmillion, Avoid the spurious concept of "relativistic mass", it will lead you astray every time.

Anyway, this question has nothing to do with mass, it's just kinematics: "a hyperbolic world line corresponds to what acceleration?" And the answer is, a hyperbola is a curve of constant curvature, hence a constant proper acceleration.

4. Sep 12, 2011

### csmcmillion

So, does that mean that I, as an observer in a spacecraft, would see an infinite constant acceleration yet never reach C?

5. Sep 12, 2011

### pervect

Staff Emeritus
As an observer in a spacecraft, you can maintain a constant proper acceleration forever, without exceeding the speed of light Note that the proper acceleration (I'm not sure if you're familiar with the term) is what you would measure with an accelerometer, or feel on the seat of your pants.

You can (and probably should) analyze the problem using only kinematics, i.e. without considering the masses or forces, initially.

The key to the kinematic analysis is a repeated application of the relativistic velocity addition equation

v_add = (v1+v2) / (1 + v1 v2 / c^2)

constant acceleration is just a procedure of constantly adding some increment to your velocity, and if you study the relativistic velocity addition formula you'll see that you will never exceed 'c' by adding velocities.

Say that after one month of acceleration, you reach .1 the speed of light relative to your speed when you started.

After another month, your speed relative to your speed one month ago is .1 c, but your speed relative to your initial speed is slightly less. Using the relativistic velocity addtion formula you find that it's only (.1 + .1) / (1 + .1*.1)

Keep repeating, and you can plot the curve of your speed vs proper time (which is the time the clock you carry with you measures).

6. Sep 12, 2011

### csmcmillion

Oh my.

Thanks!

7. Sep 12, 2011

### bahamagreen

The relativistic velocity addition equation would be used by the "stationary" observer.

The observer in the accelerating frame would not use that equation. He would measure a constant acceleration (equivalence principle) and have no reason not to integrate that to find a net relative increasing velocity (with respect to prior velocity). He can make this measure without information from outside his frame that would present results of externally viewed length contraction of time dilation (he may want to look outside and confirm the acceleration is not a local gravity source).

It spins my head, too. He can do the math and infer that after some definite time he should have passed c, because his local measure of acceleration is constant (with constant acceleration of 1 g he would infer his speed to pass c in about a year); yet any local measures of light speed will be c.

8. Sep 12, 2011

### Bill_K

I guess simulating continuous acceleration with a sequence of discrete Lorentz transformations is a good approach for someone who has never seen calculus. Me, I would rather just write down the equation of a hyperbola and differentiate it twice to see what the acceleration is!

The position vector is: x = ((c2/a) cosh (aτ/c), 0, 0, (c/a) sinh (aτ/c)) where a is the acceleration and τ is the proper time.
Note that |x| ≡ √(x12 - c2x02) = c2/a is a constant, so the world line remains a constant distance from the origin and is therefore a hyperbola.

The velocity vector is: v ≡ dx/dτ = (c sinh (aτ/c), 0, 0, cosh (aτ/c))
Note that |v| = c, as it should be.

The acceleration vector is: a ≡ dv/dτ = (a cosh (aτ/c), 0, 0, (a/c) sinh (aτ/c))
Note that |a| = a = constant, so the world line has constant acceleration a.

9. Sep 12, 2011

### PAllen

So every adult on earth should think they are moving many times c because they've felt 1 g acceleration since birth? The accelerated observer thinks they are not moving at all, just feeling force due to non-inertial path through spacetime.

What you are saying is: if I interpret the acceleration I feel relative to a hypothetical stationary observer using erroneous mathematics, I get an erroneous result.

10. Sep 12, 2011

### bahamagreen

Ok, so maybe looking out the window is not sufficient; perhaps the accelerated observer might check for presence of a gravitational gradient to distinguish between acceleration from proximity to a large mass and acceleration due to spacecraft engines.

In the later case, the accelerated observer SHOULD think he is moving. In the prior case, he should not unless he accepts the theories of Miles Mathis.

Just to enlighten me, how is the accelerating observer "using erroneous math"? In the inertial reference frame the laws of physics are expected to hold good - is it the acceleration of the reference frame that makes the math erroneous? Is there a simple way to show why the direct calculation of a change in velocity from the constant acceleration will not reach c when measured from within the accelerated reference frame?

11. Sep 12, 2011

### PAllen

1) The accelerated observer 'sees' themself at rest.

2) If they look out a window, they never see anything moving faster than c.

3) If they compute their accumulated speed relative their launch point using relativistic laws, they end up with less than c, always.

You propose, in (3) they should use Newtonian laws. Well, if you use incorrect laws you get incorrect result. Duh.

12. Sep 12, 2011

### pervect

Staff Emeritus
Only an inertial observer can integrate their acceleration over time to get their velocity in the manner you suggest. The reason for this is that the basis vectors of the accelerating observer are constantly changing.

One way to describe the correct way of getting the velocity is to say that one needs to use the "covariant derivative", not the ordinary partial derivative, but I suspect that that's not the level that we want to approach the problem at.

There's a couple of approaches we can take that may give more insight without as much sophisticated math.

Using four-vectors, as one previous poster did, you can always say that a = du/dtau, where a is the four-acceleration and u is the four-velocity, dx/dtau. We take the magnitude of the four-acceleration, we know it's direction.

You won't get the correct answer if you try to differentiate the ordinary velocity to get the four-acceleration, i.e the proper acceleration a is not dv/dtau, it is du/dtau.

The difference between u and v is that u = dx/dtau, not dx/dt, tau being the proper time of the accelerated observer, and t being the coordinate time of the inertial frame.

Less formally, we can see that if we drop off "comoving" objects at various times in our journey, at any particular instant the velocity that the comoving object measures for it's relative veocity to the origin, O, is the same velocityh that an observer at the origin, O, measures for the comoving object.

We can then use this fact to find the velocity our accelerating observer measures relative to the origin, O, at any time T, by replacing it with the velocity that the observer at the origin measures for the co-moving observer.

We can also see that the velocity measured by the comoving object will be the same as the velocity measured by the instantaneously co-located accelefrating observer.

This is why we can't integrate to get v, an it's also the reason that the ordinary velocity v does not increase without bound.

Note that the four velocity can be integrated, and does increase without bound.

13. Sep 13, 2011

### bahamagreen

Hmm, I'm thinking that an inertial observer by definition would not have any acceleration to integrate? What am I missing there?

I understand that the observer at origin and the accelerated observer would measure each others relative speeds the same magnitude and it would not reach c.

But I'm not understanding what is happening in the accelerated space craft... If the accelerated observer does not measure the relative speed of the origin but rather works solely within his local reference frame, he only has the acceleration value and no local contraction or dilation - his surroundings in the cabin are "normal". There might be some slight bending of lateral light beams and other relativistic effects, but these would be exceeding small, like at 1 G.

I don't think there would be any circumstance where the accelerated observer would conclude that the rate of acceleration was slowing down at later measures... the rate would measure constant. Within his frame of reference, must he not conclude that after some period (a year at 1 G) that these accelerations add up arithmetically?

I can see how the additions might asymptote to c because of observed time dilation from the origin. And I can see how measuring the receding speed of the origin from the accelerated craft would do the same. I'm not seeing why the accelerated observer would use those same equations locally if he observes no relativistic effects locally... missing something about that....

14. Sep 13, 2011

### PAllen

The accelerated observer sees himself stationary, just like an observer on the surface of the earth. To draw a conclusion about the effect of the acceleration, they are asking: "what does this force I feel imply about my accumulated speed relative to my starting point?". To do that, you must use some law relating this force to speed - relative to starting point. There is no speed relative to yourself. If you use the right law you get the right answer. If you use the wrong law you get nonsense.

15. Sep 13, 2011

### bahamagreen

Thanks for continuing to clarify and explain.

Am I misunderstanding what you mean by the accelerating observer seeing himself as stationary? Seems to me he would have every reason to see himself as accelerating and accumulating velocity.

The classic version of the equivalence principle places the observer within a sealed box without knowledge of whether the box is being pulled by an external cable above his head or attracted by an external mass below his feet.

But in this case, the observer has actually built or purchased a space craft with strong engines, placed it at an origin clear of gravitation bodies, activated the engines, and adjusted the throttle to provide 1 G of acceleration.

So when the observer asks himself, "What does this force I feel imply about my accumulated speed relative to my starting point?", he should think in this circumstance that he is quite justified in believing that he is accelerating and accumulating a translational velocity. I don't think he would see or feel himself as stationary.

With the known absence of a local gravitational mass and the known presence of running engines of the space craft, in order for the observer to think of himself as stationary it seems to me he would have to assume he is "holding his position" against and within an accelerating space frame that includes the origin (as if the whole universe is accelerating, but not himself) which seems Machian or etheric; he should think that is improbable.

Perhaps the problem is in the part of the question that goes, "... accumulated speed relative to my starting point?" There seem to be two ways for him to measure his speed. One involves measuring relative to the origin and using relativistic math. The other is to use the acceleration to arithmetically infer his speed. I think you are saying the former is proper and the later is improper. Is that because you say the observer would see himself a stationary?

I'm needing to understand why the accelerating observer would see himself as stationary when due to his actions with the space craft he "knows" he should be accelerating.