Differences in SR Accelerations Transformations

Discussion in 'Physics & Math' started by mikelizzi, Sep 26, 2023.

  1. mikelizzi Registered Senior Member

    Messages:
    58
    Note:
    This post uses a lot of "Tex" formatting. I never get "Tex" right the first time. My apologies if you see it mangled. I expect to be editing it in trial and error mode for a while.

    Note:
    To understand this post you need to have a knowledge of Newtonian physics, linear algebra and at least introductory level Special Relativity.

    Subject:
    I'm trying to understand the difference between two SR Transformations that apply when transforming between Inertial and Accelerating reference frames. So far this is what I have.

    The first transformation appears in two web sites. One site appears to be from the University of California Riverside.

    https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

    The two dependent equations (4) and (6) say;

    \( d = \frac{c^2}{a} \left(cosh\frac{a \tau}{c}-1 \right) \)

    \( t = \frac{c}{a}sinh\left(\frac{a\tau}{c} \right) \)

    Where:
    c = speed of light

    a = a uniform acceleration as experienced by the object and indicated by an accelerometer attached to the object (Proper Acceleration).

    (t, d) = space time coordinates of an event using the inertial reference frame.

    \( \tau \) = time coordinate of an event using the accelerating coordinate system.

    The other site is from Wikipedia.

    https://en.wikipedia.org/wiki/Acceleration_(special_relativity)

    Two two dependent equations at 6a say;
    \( x(\tau) = \frac{c^2}{\alpha} \left(cosh\frac{\alpha \tau}{c}-1 \right) \)

    \( t(\tau) = \frac{c}{\alpha}sinh\left(\frac{\alpha\tau}{c} \right) \)

    Where:
    c = speed of light

    \( \alpha \) = a uniform acceleration as experienced by the object and indicated by an accelerometer attached to the object (Proper Acceleration).

    (t, x) = space time coordinates of an event using the inertial reference frame.

    \( \tau \) = time coordinate of an event using the accelerating coordinate system.

    Note 1:
    As with the Lorentz Transformation, the coordinates x and t are really differentials. It is the distance traveled, \( x_{final} - x_{initial} \) , and elapsed time, \( t_{final} - t_{initial} \) , that is being transformed. The values of x and t alone apply only when the origins of the reference frames initially coincide, \( x_{initial} = 0 \) , \( t_{initial} = 0 \) .

    Note 2:
    When one is transforming position and time coordinates with respect to an accelerating reference frame to coordinates with respect to an inertial reference frame (this case), the time component of the first set of coordinates, \( \tau \) , is called Proper Time and the time component of the second set, t, is called Coordinate Time. Proper Time is the time recorded on the accelerating object's clock. It is frame invariant. Coordinate Time is the time component of the space-time coordinates of the event with respect to the observer (the inertial object in this case). It is not frame invariant. If one flips the transformation then the assignment of which is the Proper Time and which is the Coordinate Time also flips.

    Note 3:
    When describing the motion of a point object, there are two kinds of acceleration in relativity. Proper Acceleration was just defined above. There is also Coordinate Acceleration, which is defined in the traditional way as dv/dt, the change in the observed velocity of an object divided by the change in that observer's time. Proper Acceleration is frame invariant. Coordinate Acceleration is not. These transformations, and this discussion, applies to problems where Proper Acceleration is given and is constant. This discussion does not apply to problems where Coordinate Acceleration is given and is constant. (An analogy would be that there are two kinds of length in relativity, Proper Length and Coordinate Length.) When an accelerating object's velocity is momentarily 0 relative to some inertial frame, the Coordinate Acceleration relative to that frame is equal to the Proper Acceleration. For other circumstances, if one value is known, the other can be calculated.

    The two transformations are the same. They just use different notation. Actually I'm not sure they can be called transformations. They don't map the coordinates of time and position between two reference frames. I don't see time and position with respect to an accelerating reference frame mapping to time and position with respect to an inertial reference frame. What I see is time with respect to an accelerating reference frame mapping to time and position with respect to an inertial reference frame. If these equations are a transformation, they are not generic. They are for the special case where the position of the object with respect to the accelerating reference frame is 0. If that is true, it has important implications when applying these transformations to the traditional Bell's Spaceships problem, the one with two rockets. One rocket can be declared to start at position 0 with respect to the accelerating reference frame. OK. But then the other must start with a nonzero value for position. It must have an offset and neither version of the above transformation applies.

    Compare the above to a more generic transformation from the following textbook.

    Basic Relativity (Chapter 8)
    By Richard A. Mould
    Copyright 1994 Springer-Verlag, New York, NY

    This textbook says (using my personal notation);
    \( x(t_{accel}, x_{accel}) = \frac{-c^{2}}{a_{proper}} + \left(x_{accel} + \frac{c^2}{a_{proper}}\right)cosh\left(\frac{a_{proper}ct_{accel}}{c^2}\right) \)

    \( ct(t_{accel}, x_{accel}) = \left(x_{accel} + \frac{c^2}{a_{proper}}\right)sinh\left(\frac{a_{proper}ct_{accel}}{c^2}\right) \)

    Where:
    c = speed of light

    \( a_{proper}\) = a uniform acceleration as experienced by the object and indicated by an accelerometer attached to the object (Proper Acceleration).

    (ct, x) = space time coordinates of an event using the inertial reference frame.

    \( (ct_{accel}, x_{accel}) \) = space time coordinates of an event using the accelerating coordinate system.


    Note 4:
    \( x_{accel} \) in the above transformation is, I think, the distance to the accelerating object using the accelerating coordinate system, \( S_{accel} \) . The object is at rest using that coordinate system so \( x_{accel} \) does not change with time. \( x_{accel} \) may be thought of as the offset in \( S_{accel} \) . In the case where the origin of the accelerating coordinate system is initially at rest and coincides with the origin of the inertial reference frame, S, then \( x_{accel} \) , is also the initial distance using the coordinates of S (meaning initially, \( x_{accel} = x \) ).

    A comparison to the Lorentz Transformation
    Recall that the Lorentz Transformation applies between two inertial reference frames.
    \( x'(t,x) = \gamma(x -\beta ct) \)

    \( ct'(ct,x) = \gamma(ct-\beta x) \)

    Where:
    \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \)

    \( \beta = v/c \)

    (ct, x) = space time coordinates of an event using one of the inertial reference frames

    (ct', x') = space time coordinates of an event using the other inertial reference frame

    Note that the above transformation is generic. You can transform any space time coordinates (x, ct) from the unprimed reference frame to the primed reference frame (x', ct').

    Now consider the following transformation.
    \( x'(t) = \gamma(-\beta ct) \)

    \( ct'(ct) = \gamma(ct) \)

    I expect you would immediately recognize that the above transformation is a special case of the Lorentz Transformation, one where x = 0. Similarly, I believe the relativistic transformation from Wikipedia and the ucr web sites are copied from a derivation where the initial conditions are that the two reference frames coincide and the accelerating object is at that coincident point. I know the relativistic transformation derived by Richard Mould allows the accelerating object to have an initial offset because the derivation is provided in the textbook. (All require the initial velocity to be zero.)

    I'm going to end my initial post here. I've covered a lot of material and I'm not 100% sure I have it all correct. If anyone has knowledge of these transformations and sees a mistake please comment.

    P.S.
    I'd like to keep this discussion free of space-time diagrams. Space-time diagrams are visual aids. The way they are drawn and the way they are interpreted is determined by the math they are intended to describe. My experience with space-time diagrams is that people get bogged down arguing about how to draw them and how to interpret them and never get to the source math.
     
    Last edited: Sep 26, 2023
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  3. Halc Registered Senior Member

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    Jut a couple initial comments.
    Two slightly different symbols for the same thing. Proper acceleration is frame independent so a different symbol is not needed for inertial frames and the accelerating frame respectively.

    Notice that most sites (wiki say) use the word 'particle' instead of 'object' since the latter implies extension, an object with dimensions, but since both proper time τ and proper acceleration α differ at different displacements from where 'the object' is considered to be, it's best to use 'particle' which emphasizes the fact that the value only applies to what is essentially the origin of the accelerating reference frame.
     
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  5. mikelizzi Registered Senior Member

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    Thanks,

    Using the term "particle" would be consistent with the rest of the world and less subject to misinterpretation. I will use particle from now on. If you are referring to the use of Greek alpha for acceleration rather than "a", it did seem strange that Wiki would use it. I will avoid using it.
     
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  7. Neddy Bate Valued Senior Member

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    2,547
    mikelizzi,

    Referring to your point about the equations requiring the particle to be located at the origin of its own (accelerating) reference frame's coordinate system, and your mention of Bell's scenario:

    In Bell's scenario, it is given that both rockets have the same coordinate acceleration in the unprimed frame, and from that, we easily conclude that the distance between them (center-to-center) is therefore constant in that frame. From this fact, we can conclude that both rockets have identical accelerometer readings. The very definition of an accelerometer being a device that measures acceleration requires that. Equal coordinate acceleration of two rockets as measured in an inertial frame must mean equal proper acceleration assuming normal physics assumptions.

    From the coordinates in the unprimed frame, you should be able to reconstruct what happens in the primed frames when there are two particles of identical acceleration which are not both at the origin. You just have to be careful to transform everything correctly.
     
  8. James R Just this guy, you know? Staff Member

    Messages:
    39,277
    mikelizzi:
    As you suggest later, this is a special case, not a general transformation of coordinates. Specifically, these equations relate to a scenario where the proper acceleration is constant. But, then again, the Lorentz transformations also relate to a specific scenario, in which two inertial frames have a constant relative velocity.

    I, personally, would not use the word "differential" to describe this.

    You are correct, however, in that if we put $\tau=0$, then these equations give $d=x(\tau)=0$ and $t=0$. That is, the assumption is that the origins of the spatial coordinates are co-located when $t=0$ and $\tau=0$.
    No. The proper time does not "flip" if you invert the transformation. A proper time interval between two events in spacetime is measured in a reference frame in which the two events occur at the same spatial coordinate. All frames agree on who (if anyone) is measuring a proper time interval between any two specified spacetime events. Proper time intervals are not frame dependent.

    In the special case we're considering here, in which has one of the frames has constant proper acceleration, any clock that is stationary in the accelerating frame will measure proper time.
    Correct.
    A proper length is the length of an extended object, measured in a frame in which that object is stationary in the spatial coordinates. I don't think the term "coordinate length" is very helpful; the question is: does a given exended object have a proper length in some reference frame or other? If it does, then all observers in every frame must agree on what the proper length is.
    Yes, the two transformations are the same. However, they do map the coordinates of time and position between two reference frames - one being an inertial frame and the other being a frame with constant proper acceleration.

    It is important to realise what is actually being "mapped" or "transformed" by these equations (and by the Lorentz transformations of the spacetime coordinates). Fundamentally, these equations transform the spacetime coordinates of events in spacetime. An event is anything that happens at a single point in spacetime. You can think of events as being fixed points in spacetime. Events either happen or they don't happen. If they happen, then all observers agree that they happen, and similarly if they don't happen. Observers in different frames, however, can assign different spacetime coordinates to the same event.
    The Lorentz transformations also "mix" the space and time coordinates together. Realising this leads to an understanding that spacetime is not really "space + time", considered separately. Rather, what an observer calls "space" or "time" depends on his or her frame of reference.
    The Lorentz transformations assume that the primed frame is travelling at constant velocity in the positive $x$ direction. So, in a sense, they are no more "generic" than the transformations that assume that the primed frame is accelerating in the unprimed coordinates. Obviously, because the two frames we are concerned with are moving in different ways relative to one another in the two scenarios, the applicable transformations will be different.
     
    Last edited: Sep 27, 2023
  9. mikelizzi Registered Senior Member

    Messages:
    58

    Whoops. You are right about the initial conditions of Bell's Space Ships. I just checked Wikipedia and it agrees with you that the initial conditions are that the Coordinate Acceleration is the same for both ships. Good thing I didn't make any calculations yet.

    However, I disagree with your conclusion that their Proper Acceleration (accelerometer reading) is the same throughout.
    " Equal coordinate acceleration of two rockets as measured in an inertial frame must mean equal proper acceleration assuming normal physics assumptions." That is true only at the start of the problem, at time zero, when both rockets have relative velocity zero. After that it is just the opposite. Equal Coordinate Acceleration for the 2 rockets (with respect to the initial inertial observer) requires them to have different Proper Acceleration, and visa-versa. That is one of the consequences of the transformation that includes an offset.

    You can verify this very non-intuitive consequence yourself. Use the transformation that allows for an offset. It is written to transform the values of a rocket's Proper Acceleration, Proper Time and initial distance to distance and time values with respect to an inertial observer. (I know that is not the initial conditions of the problem. Please be patient.) Make two calculations. Use the same Proper Time, same Proper Acceleration but different initial separation for the 2 rockets. You will need some large values. Relativistic effects, as usual, only show up for large distances and velocities. Look at the resulting distances. The separation of the rockets with respect to the inertial reference frame will be different. The only way that can happen is if they had different accelerations with respect to that inertial observer. That means different Coordinate Accelerations for the same Proper Acceleration. If you require the Coordinate Accelerations to be the same for the 2 rockets, the initial conditions of Bell's Space Ships, you will need to assign different Proper Accelerations. That is a direct consequence of the offset in the transformation. If you use the transformation commonly presented on the internet, one without an offset, you are transforming values for 2 spaceships where both start at the origin. Adding the separation after you make the transformation will, of course, give you a constant separation. But you are not making a valid transformation.

    Having typed all that, I don't expect such a long winded response to persuade you. But it's the best I can do.
     
  10. Halc Registered Senior Member

    Messages:
    350
    Apologies that I have more comments to James' reply than to the OP
    A specific scenario between inertial frames, period. It is not possible for two inertial frames to not have constant velocity between them.

    But we're not measuring the frame independent interval between two events here, but rather the frame independent length of a curved worldline connecting the two events, which mikelizzi correctly says is the time recorded on the accelerating clock. It is computed not by taking the interval between the two end events, but by integrating the interval along the clock path.
    Perhaps I misread the case to which you (James) are responding since part of Note 2 also referred to the inertial case.

    They also agree on the proper times measured on multiple non-inertial worldlines. I mean, the whole point of differential aging (the twins scenario) is different measurements of proper time between the same two specified spacetime events. There is but the one interval, but many different proper lengths of different paths between those two events.

    Again, true even outside this special case. Clocks always measure proper time by definition.

    Agree, but in a more general case, if the object is not everywhere stationary at a given time, it's proper length is still determined by integrating along its length, computing each bit in a different frame in which the bit is stationary. So for instance, the proper length of a rotating circle of wire can be computed despite the lack of inertial frame in which the wire is stationary. In the inertial frame of the center of the loop, not one 'bit' of it is stationary.

    So in light of the above, everybody must agree on that proper length. Proper length being frame independent, the part I bolded above is worded wrong, and should instead read "does a given exended object's length in some reference frame match its proper length?" It is not possible for the object to not have a proper length.

    In physics, an event IS a point in spacetime, something to which coordinates can be assigned. Whether anything 'happens' there or not is irrelevant. So I balk at the bit about them 'happening or not'. Events exist or not. For instance, an absolutist might deny the existence of an event inside a black hole since they lack coordinates. As such, they deny the existence of black holes altogether.
     
  11. mikelizzi Registered Senior Member

    Messages:
    58
    I guess I didn't get my point across. Yes, the transformations are for constant Proper Acceleration. But that is not the reason I think they are not generic. The reason is that the first 2 transformations DO NOT transform any time and any position from an accelerating reference frame to an inertial reference frame. They transform any time and ONE position (x=0) from an accelerating reference frame to an inertial reference frame. The third transformation, the one from the Mould textbook transforms any time and any position from an accelerating reference frame to an inertial reference frame.

    Yep. I really didn't want to write that. Wish I knew of a more appropriate word.

    Bad choice of words. Of course I agree with what you typed.


    Again, I guess I didn't get my point across. The first 2 transformations DO NOT transform any time and any position from an accelerating reference frame to an inertial reference frame. They transform any time and ONE position (x=0) from an accelerating reference frame to an inertial reference frame. The third transformation, the one from the Mould textbook transforms any time and any position from an accelerating reference frame to an inertial reference frame.
     
  12. Halc Registered Senior Member

    Messages:
    350
    No, it is true always, presuming equal initial velocity in that inertial frame (call it S), something I don't see stated here. But if initial velocity and acceleration are identical, then velocity must remain identical at all times in that inertial frame.

    So I very much disagree with this.

    Which transformation is this? Most of them are of the form v = (function only of initial v, α and τ or t). It is not a function of offset (d), so the velocity relative to S after any amount of time as measured by either S or the accelerating object must be identical.

    I did look at the distances, and they're not different than they were at the beginning. Coordinate acceleration must be the same if both v (in S) and α are the same.

    This is wrong. You've not shown it, except by incorrectly assuming that the distance in S changes, which it doesn't.

    It would help if you quoted exactly which transformation you're using to conclude all this. It cannot be right if the above results.
     
  13. mikelizzi Registered Senior Member

    Messages:
    58
    I'm not referring to a velocity transformation here, though I suppose one is implied. I am referring to the transformation between an accelerating and an inertial reference frame where the reference frames initially coincide and the initial relative velocity is zero, this transformation from the initial post.

    \( x(t_{accel}, x_{accel}) = \frac{-c^{2}}{a_{proper}} + \left(x_{accel} + \frac{c^2}{a_{proper}}\right)cosh\left(\frac{a_{proper}ct_{accel}}{c^2}\right) \)

    \( ct(t_{accel}, x_{accel}) = \left(x_{accel} + \frac{c^2}{a_{proper}}\right)sinh\left(\frac{a_{proper}ct_{accel}}{c^2}\right) \)

    Where:
    c = speed of light
    \( a_{proper} \) = a uniform acceleration as experienced by the object and indicated by an accelerometer attached to the object (Proper Acceleration).
    (ct, x) = space time coordinates of an event using the inertial reference frame.
    \( (ct_{accel}, x_{accel}) \) = space time coordinates of an event using the accelerating coordinate system.

    That transformation tells me 2 particles that have an initial separation with respect to an inertial observer will not keep the same separation even though they have the same Proper Acceleration. My conclusions follow from that. Am I reading the transformation wrong?
     
  14. Halc Registered Senior Member

    Messages:
    350
    OK, I didn't really understand that bit, hence my continued confusion. Again, say frame S is the inertial one, and R is the acceleRating one, just for ease of notation. If there's a second accelerating particle, it's accelerating frame can be represented with R'.
    You say these equations compute x and t in S from coordinates given in R. Is that right?
    Somehow, from this, you seem to conclude that some of v, coordinate acceleration, proper acceleration, and separation of the two particles in S are at any moment in time different.

    The bold part lacks a frame reference. You reference their initial separation at t=0 when all three frames have the same separation, but not which frame is measuring the subsequent separation.
    They will very much maintain their constant initial separation in S, and the above doesn't contradict that. They will not maintain constant separation in either R or R', but you didn't say that.

    The above equations assume the accelerated frame shares the same origin as the inertial frame, but if not, it is a simple matter of adding the initial displacement to the final result.

    Maybe. Let me simplify it. If we use natural units, any reference to c disappears. If we set proper acceleration to 1, that also vanishes.
    How are you using x_accel here? The accelerating particle is always at the origin of its own accelerating frame, so x_accel is zero.
    We're accelerating point objects here, so all these terms cancel out, leaving just τ, which reduces the above general equations to simply:

    \( x(t_{accel},d_{initial}) = cosh\left(t_{accel}\right) - 1 + d_{initial} \)

    \( t(t_{accel}) = sinh\left(t_{accel}\right) \)

    Note that d_initial is the location in S of the particle accelerating at '1', not the displacement in R from said particle.
    The above shows that x is the same for both particles, except for the initial displacement.
     
    Last edited: Sep 27, 2023
  15. Neddy Bate Valued Senior Member

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    2,547
    I agree with Halc that you must be making a mistake someplace if you are concluding that.
     
  16. mikelizzi Registered Senior Member

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    58
    I going to reply one part at a time.

    I am not concluding that Proper Acceleration is different. Same Proper Acceleration for the 2 particles is a given. But I am concluding that clock rate is different for the 2 particles. Different clock rate applied to same Proper Acceleration gives different Coordinate Acceleration, different Velocity and different Distance traveled.

    By clock rate I mean dt'/dt where dt is the elapsed time for the inertial observer and dt' is the elapsed time for the accelerating particle. What I have concluded is that, if 2 particles are at rest at different locations with respect to the accelerating reference frame they will have different clock rates with respect to an inertial observer. That is NOT what happens between inertial reference frames. As you know, all particles at rest with respect to one inertial reference frame have the same clock rate relative to another inertial reference frame, no matter their position relative to that first reference frame. I think that is part of the definition of a reference frame. Perhaps that is why I have seen it written that there is no such thing as an accelerating reference frame. And that may be why the author to whom I refer, Richard Mould, uses the phrase an "accelerating system of coordinates".

    His textbook uses about 3 pages to explain this. I will not try here.

    I'm going to stop here and not discuss your other comments because I am certain you will object the the above conclusion. It's not my conclusion. I'm just borrowing it.
    Would it be legal for me to scan a few pages of a copyrighted book and post them as jpg files, assuming I know how to do that?
     
  17. Mike_Fontenot Registered Senior Member

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    622
    Can you provide a reference to his textbook that you are talking about?
     
  18. Halc Registered Senior Member

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    350
    Clocks by definition run at one second per second. You probably don't mean anything like that, but your statement isn't entirely clear. You seem to be referencing their rate in S, the initial inertial frame.
    Two synced clocks in S simultaneously commencing identical acceleration will remain in sync in S indefinitely. By relativity of simultaneity, that means that they will not be in sync relative to pretty much any other frame. So the clock rates are the same in S, even if they're not the same as a clock stationary in S.
    This is pretty trivial to demonstrate since the clocks always have the same coordinate speed in S, and thus the same dilation relative to S.

    I suppose so, but the clocks rates relative to S are necessarily identical, per the logic above, and per the equations you quoted.

    OK, that's a frame reference to S. That should have been stated right up front, but at least it's there. Yes, dt'/dt is identical for both particles since the speed in S is always identical at all times (dt).
    But below, you switch to a different frame:

    Your two accelerating particles are never both at rest in the same frame at the same time in that frame. They have identical velocity at a given moment in S. By relativity of simultaneity, that means that they don't have identical velocity relative to a different frame. You are perhaps speaking of two particles at opposite ends of a long rigid object, instead of the situation of Bell's scenario with two independently accelerating particles with identical proper acceleration. The opposite ends of a long accelerating rigid object will experience different proper accelerations. Both ends will be stationary in the frame of hte acceleration object, but indeed the clock rates will be different in that frame, as they are different in any frame except S in Bell's scenario.

    But nothing is inertial in this scenaio. For clocks to have the same rate, they need to maintain identical speed over time in that frame, not just be momentarily stationary in it.

    I suppose it depends on what offends you, but one can very much specify the frame of say a long rocket, or at least the frame of a particular location in that object. Non-inertial frames lack many of the properties of inertial ones, such as Newton's basic laws, energy/momentum conservation, etc.

    What exactly are you trying to convey? You're all over the map with the scenarios, with one acceleating particle, then two with identical acceleration, and then two different particles that remain stationary relative to each other. These are all very different scenarios, each with different mathematics and different transformations. The formula you quoted in your last post seem to be for the 3rd case, the rigid object case, the only one where x_accel is a meaningful value.

    Not my call, but I see people do it quite often, so long as it's not for the purpose of selling an education. Don't copy the whole book, just what's needed for the discussion.
     
  19. mikelizzi Registered Senior Member

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    Never mind. I don't know what you are talking about and I believe you don't know what I am talking about. I see no point in continuing this thread.
     

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