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A rotating Hot Dog about an inertial observer

  1. Jul 18, 2014 #1
    According to Einstein (e.g. in his book The Meaning of Relativity), a clock rotating about a central clock will be judged by the central clock to run slower than the central clock.

    This means that a signal sent by the central clock will be perceived by the rotating clock as being of a higher frequency.

    If the central clock sends a very low frequency radio signal with a wavelength of 32000 metres, then at the right linear speed for the rotating clock it receives this same wave as only 0.12 metres in length, which is a microwave.

    Now a microwave would leave a telltale mark on the rotating clock! OK...here comes the hot dog. To confirm that waves leave visible marks, you can try a home experiment. You remove the turning plate from a microwave oven and place a hot dog inside the oven. After 2 minutes or so some bubbling will appear on the hot dog; the total length of the bubbling will be about 0.12 metres, confirming the length of one microwave.

    Therefore, in the frame of the rotating clock a meter stick would measure each wave as contracted compared to the waves received in an inertial frame. So far so good...but....

    As each wave sent from the inertial central position was 32000 metres in length, and now the wave mark on the rotating clock ( or the hot dog if you prefer) is only 0.12 metres, then, to an idiot, the rotating clock has expanded in relation to that one wave. Equally, we may say the wave has contracted in relation to the rotating clock.

    You know the wave-mark on a hot dog is real, right?

    So, in view of the principle of length contraction, which I do not dispute, what is the “fabric” that the “expanding” meter stick in the rotating frame is measuring? Is it a worldline in Minkowski space? Is it direct evidence of length contraction itself? What should we call this metric/dimension/property/whatever ? When I asked this problem elsewhere I was lectured "moving bodies contract, not expand". Yep. I know that. But this is a different perspective altogether.

    I trust someone has actually thought of this problem before, or can point to an example in a reference.
     
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  3. Jul 18, 2014 #2

    TumblingDice

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    Correct. EDIT: The rotating clock will also judge the central clock to run slower than itself. This is a key part of relativity. The clocks/observers will maintain these observations as long as they are separated in their relative motions.

    Following your example, this observation should indicate that the rotating clock perceives signals from the central clock in a lower frequency (time is observed as the central clock running slower).

    I think this as a good place to pause for reflection, establish a proper foundation, and wait for further questions before moving forward. :smile:
     
    Last edited: Jul 18, 2014
  4. Jul 18, 2014 #3

    WannabeNewton

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    Yes this is true. It will be given by ##\frac{\nu}{\nu_0} = \gamma = (1 - \omega^2 R^2)^{-1/2} > 1## where ##\omega## is the angular velocity of rotation of the rotating clock about the central clock. This is very easy to derive in the frame corotating with the rotating clock. In this frame the central clock and the corotating clock both follow orbits of the time-like Killing field ##\xi^{\mu}## i.e. the 4-velocity field of the family of rotating clocks, which includes the degenerate case of the central clock, is ##u^{\mu} = \gamma \xi^{\mu}##. If a light signal is sent from the central clock to a rotating clock at ##R## with the associated null geodesic having wave 4-vector ##k^{\mu}## then the frequency of the light signal as measured by the central clock will be ##\nu_0 = -\xi^{\mu}k_{\mu}|_{r = 0}## whereas that measured by the rotating clock will be ##\nu = -\gamma \xi^{\mu}k_{\mu}|_{r = R}##. But ##\xi^{\mu}## is a time-like Killing field so ##k^{\nu}\nabla_{\nu}(k^{\mu}\xi_{\mu}) = 0## i.e. ##\frac{\nu}{\nu_0} = \gamma = (1 - \omega^2 R^2)^{-1/2} > 1##. This is of course just the Doppler blueshift.

    I couldn't fully follow your description because it uses quite many words imprecisely, making it confusing to figure out exactly what you are trying to describe; however, I'll try to comment on what I could make of it. The wave-mark pattern on the hot dog will be frame-dependent. Barring philosophical discussions about what "real" is, any measurement of the wave-mark patter will involve simultaneity. But simultaneity is relative. Certainly the simultaneity of a moving observer relative to that of a stationary observer will almost everywhere disagree. This means that there is no reason to expect the wave-mark pattern on the hot dog relative to one observer to be the same as that relative to another moving observer because of the relativity of simultaneity. This is without even touching the subtleties involved in defining simultaneity for a rotating clock.

    Furthermore, it makes no sense to say that the rotating clock has "expanded" in relation to the light signal, or relative to the light signal. How would you characterize that mathematically and, more importantly, operationally? You can't even define the rest frame of a light signal let alone measure lengths relative to it. This isn't even taking into account the fact that clocks in relativity are idealized as points so they don't have any length capable of expanding or contracting. Yes the light signals have smaller wave-lengths when measured by the rotating clock when compared with the larger wave-lengths measured by the central clock but in absolutely no way does this imply that the rotating clock has "expanded" in relation to the light signal for the reasons mentioned.

    There is no expanding meter stick arising from purely relativistic effects in the rotating frame.
     
  5. Jul 18, 2014 #4

    TumblingDice

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    Thank you, WbN! Ya' know, I was thinking of uniform circular motion when I replied, and erred on the wrong side (trying to keep it out of the smoke.) I've had an uneasy feeling about that and just checked back in. All my bad for replying too quickly with a ready-shoot-aim approach.

    To make sure my feet are on the ground... Would it be O.K. to apply an extension of the equivalence principle as a perspective into this. IOW, assuming enough distance from the central clock, the rotating observer can take the view that they're uniform acceleration is gravity, and they're viewing the central clock in a fixed position directly overhead? Hence it's a gravitational well that precipitates the blueshift...
     
  6. Jul 18, 2014 #5

    WannabeNewton

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    This is uniform circular motion, isn't it? One clock is in uniform circular motion around the other clock.

    Yes you can think of it that way. In fact that's exactly what I did in my calculation above.
     
  7. Jul 18, 2014 #6
    Let's say the rest lenght of the hot dog is 1.2 m. This hot dog will contract to a five micrometers long hot dog in our example, if I have calculated correctly.

    If we want distinct marks on the hot dog, we need uneven heating, so we need a standing wave. So let's say a metal fork is stuck at the end of the hot dog. Let's say the waves reflected from the fork and the waves straight from the emitter cancel out at the middle of the hot dog.

    Now here is a question: How can a crest and a trough fit into 5 µm long resonator chamber? Answer: Each wave stays a long time inside this resonator chamber, because the resonator chamber is moving so fast. It's a time dilated light clock.
     
    Last edited: Jul 18, 2014
  8. Jul 20, 2014 #7

    Dale

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    Closed pending moderation. EDIT: the thread will be reopened, and I remind the participants to avoid personal speculation.
     
    Last edited: Jul 20, 2014
  9. Jul 20, 2014 #8
    Thanks jartsa, WannabeNewton and TumblingDice

    My first reply to your question was moderated, unfortunately due to imparting my own personal interpretation to it. So I must be more careful. I think we could safely characterize the "expansion" as an an expansion in time. In this case, only the sketch expands, not the object.

    I agree, that is correct. The crest and trough stay a longer time in the time-expanded resonator chamber. This means a body will have more wave-marks in the rotating frame than it would in an inertial frame. This was also pointed out by WannabeeNewton.

    It is easy to make mistakes about frequency! I myself have often erred and thought "lower frequency equals slower time". We must be careful on this point. The frequency of signals (sent from an inertial frame) as received in a non-inertial frame is not the same as the frequency that a clock ticks over in that same frame.

    WbN, (if I may call you that), this is very interesting. Does this mean the central clock receives frequencies that are the reciprocal of the Doppler blueshift, so that the central clock receives a lower frequency (or Doppler redshift)?
     
  10. Jul 20, 2014 #9

    WannabeNewton

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    Yes, light signals sent from the rotating clock to the central clock will be doppler redshifted to the central clock. An easy way to immediately see this is from the fact that the above calculation has time translation symmetry so I can just as well run it backwards in time and it will be entirely equivalent to a null geodesic emanating from the rotating clock to the central clock. Therefore the formula ##\frac{\nu}{\nu_0} = \gamma## still holds but with the interpretation now that it is the ratio of the frequency ##\nu## of the light signal emitted and measured by the rotating clock to that measured and received by the central clock, ##\nu_0##.

    You can think of this, if you wish, through the equivalence principle. We can imagine the rotating clock, as well as the central clock, as being at rest in a gravitational field with a gravitational potential which is given more or less up to a logarithm by ##(1 - \omega^2 r^2)^{1/2}##. You'll notice that this potential decreases with ##r##. Therefore if a photon is sent from the central clock to the rotating clock, it is equivalent to a photon going from a higher potential to a lower potential in this gravitational field. In doing so it gains energy and as a result ends up with a higher frequency at the location of the rotating clock, which is at rest in this gravitational field. Conversely, a photon sent from the rotating clock to the central clock is going from lower potential to higher potential so it loses energy because it is "climbing out" of the gravitational field.

    Of course there is nothing mysterious going on here and we can analyze this using just special relativity. First imagine you are at the central clock and you have aligned a photon gun in some direction, call it the ##y## axis. There is of course a clock rotating around you at some fixed radius ##R## with angular velocity ##\omega##. You time the shot of your photon gun so as to arrive at the rotating clock right when it passes by the ##y## axis. For example say the clock is initially, that is for ##t = 0##, at ##\phi = 0## then it will take it a time ##t = \frac{\pi}{2 \omega}## to reach the ##y## axis relative to the synchronized clocks in your rest frame and if you emit a photon at time ##t_0## it will take it a time ##t = t_0 + \frac{R}{c}## to reach the desired point; we want these two times to coincide so ##t_0 = \frac{\pi}{2 \omega} - \frac{R}{c}##.

    With this in place, the clock will at the instant it passes the ##y## axis be traveling in a direction perpendicular to it so we can call the instantaneous axis of motion of the clock the ##x## axis. The direction of the photon is given by ##\vec{n} = \hat{y}## so its wave 4-vector in your frame is ##k^{\mu} = \nu_0(\hat{t} + \hat{y})##. We then Lorentz boost in the ##x## direction with speed ##v = \omega R## to the inertial frame momentarily comoving with the rotating clock at this instant. Then ##\nu = \Lambda^{t'}{}{}_{\mu}k^{\mu} = \Lambda^{t'}{}{}_t k^t + \Lambda^{t'}{}{}_{x} k^x = \nu_0 \gamma## i.e. ##\frac{\nu}{\nu_0} = \gamma > 1## as we already had. This is known as the transverse doppler shift.

    Conversely, imagine that when the rotating clock passes the ##y## axis it emits a photon towards the central clock. Obviously in the rest frame of the central clock the photon will just travel in a straight line along the ##y## axis towards the central clock. But in the inertial frame momentarily comoving with the rotating clock at this instant, this emitted photon will not be aligned with the ##y'## axis because of the relative speed ##v = \omega R## the clock has at this instant in rotating around the central clock; it will have some angle ##\phi'## measured from the ##x'## axis. Let the frequency of the photon in the rest frame of the rotating clock be ##\nu'## and that measured by the central clock ##\nu'_0##. Using the inverse Lorentz transformation, we have ##\nu'_0 = \gamma \nu'(1 + v\cos\phi')##. Furthermore, again through the inverse Lorentz transformation, we have ##\cos\phi = \frac{\cos\phi' + v}{1 + v\cos\theta} = 0## so ##\cos\phi' = -v## hence ##\nu'_0 = \gamma \nu'(1 - v^2) = \gamma(\gamma^{-2})\nu' = \gamma^{-1}\nu'## which is of course not surprising in the least.
     
  11. Jul 20, 2014 #10

    Dale

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    One thing that I think is important to mention in this discussion is the fact that the standard length contraction formula applies only in certain circumstances. Specifically, the length must be constant, the length must be at rest in one of the frames, and both frames must be inertial. Light is never at rest in any frame, so the standard length contraction formula never applies to light.

    Also, in the hot-dog scenario is not the same as a microwave oven. In a microwave oven the microwaves form standing waves where the nodes cause the hot spots and the antinodes cause the cold spots. A central radio emitter would broadcast a travelling wave, not a standing wave. There would be no hot or cold spots.

    However, the frequency and wavelength could certainly be measured regardless, with no need to reference a hot dog. However, again, the usual length contraction formula would not apply for two reasons, first because light is not at rest in either of the frames and second because the rotating frame is non-inertial.
     
  12. Jul 21, 2014 #11
    That was a great answer, WbN. I would be grateful if you could advise a textbook where this issue is discussed. I have several texts, including one on Mathematical Physics, but the question was not answered in these.

    A few questions arose in my mind from your comprehensive reply.

    I understand a photon may be said to have zero rest mass but it has positive relativistic mass. Then, according to [itex]e=mc^2[/itex]does the energy gained by the photon as it goes from a higher potential to a lower potential correspond to an increase of relativistic mass? Conversely, a body’s relativistic mass would need to decrease as it goes from a lower potential to a higher potential in the gravitational field. After all this, can we say a rotating clock’s relativistic mass increases relative to a central clock?

    You showed how special relativity may explain the slowing of time for a non-inertial clock by using an inertial frame momentarily comoving with the rotating clock at a particular instant. You did this by pointing a photon gun that is timed to send a photon to the rotating clock as it reached the [itex]y[/itex] axis with an inertial comover at the same instant.

    It seems necessary that the rotating clock receives the photon perpendicular to the direction of motion. That idea, I think, comes from the light-clock (double mirror system) for inertial frames in special relativity.

    What if we did away with the inertial comover? I am wondering if your conclusion could also be derived with a large number of rotating clocks (light-clocks) following each other. One of these rotating clocks could emit a photon from a mirror. Since it is a rotation, the photon would miss the second mirror facing parallel in the same light-clock. (Because in the rotating frame, a photon will appear to bend away, not go “vertically” as in an inertial system). However the photon would enter the light-clock system of a different clock system that is following it in the rotation. The path traced by the photon, in relation to 2 rotating light-clocks, will not be an obvious double diagonal- it will be a little curved. However, if the linear distance (i.e. the distance of the curve in the rotation) between each mirror for each of the separate light-clocks was translated to an x, y coordinate system (that is, into flat geometry) and the flat distance between each struck mirror was presented as a straight line for the path between the mirrors (even if they belong to separate light-clocks) would not the familiar right triangle appear? This may be a simpler way to visualise the Lorentz transform, even though we have cheated by using more than one light-clock.

    Yes, I guess it can still be technically possible if the rotation takes place in an oven that makes a standing wave...but oh the technical difficulty with that!

    As light has no rest frame I agree the length contraction formula does not apply to light, according to the standard description of special relativity theory.
     
    Last edited: Jul 21, 2014
  13. Jul 21, 2014 #12

    WannabeNewton

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    This should be discussed in most if not all SR textbooks. My favorite is Rindler: https://www.amazon.com/Relativity-Special-Cosmological-Wolfgang-Rindler/dp/0198567324

    The concept of the relativistic mass is a poor one and definitely an eschewed one. You should not think of the photon as having a relativistic mass. It simply has an energy ##E = \hbar \omega##.

    In the scenario you described there is no other possibility. The central clock is inertial so any light signal emitted by it will travel in a straight line through its inertial frame. Any such line emanating from the central clock will necessarily intersect the rotating clock perpendicular to its instantaneous direction of motion because it is traveling in a circle about the central clock.

    Are you talking about a family of clocks all rotating about the central clock with the exact same angular velocity? If so then you can readily apply the calculation in post #3 to find the Doppler shift between two different clocks in the family.
     
    Last edited by a moderator: May 6, 2017
  14. Jul 21, 2014 #13

    A.T.

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    What about aberration?
     
  15. Jul 21, 2014 #14

    WannabeNewton

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    What about it? We're talking about the direction in which a light signal emitted by the central clock arrives at the rotating clock in the inertial frame of the central clock. This is clearly always perpendicular to the direction of motion of the rotating clock which is tantamount to saying that the tangent to a circle is always perpendicular to its radius; put more concisely, the light source is the central clock so there is certainly no aberration in this frame of light signals emitted by it, which is what is being discussed here.

    There is of course aberration of the light signal upon arrival at the rotating clock in the inertial frame momentarily comoving with it at that instant but that is not what is being discussed here. For that see the last paragraph of post #9.
     
  16. Jul 21, 2014 #15
    Yes, the clocks rotate with the same angular velocity. But they are light clocks. I should have been more precise in post #11. There is a technical problem in shooting a photon from the central clock into this family of light clocks. To get around that, the central clock could send a signal to one of the light clocks that has a sensor. When the signal strikes the sensor a separate “photon gun” within that light clock releases a photon, and the photon travels along the y axis from the perspective of the central clock. Next, this photon escapes the light clock due to the rotation. However, one of the mirrors in the next rotating light clock in the family is struck by the same photon as it meets the same point on the y axis. The two light clock systems, taken together, can be translated onto a plane to eliminate the curved path so that they resemble a single light clock in inertial motion. So in effect the path of the photon emitted from within the first light clock can be represented as a diagonal. From there, the usual length contraction and time factors can be seen.

    There is one other issue, WbN, that may not have been clarified, which was raised by TumblingDice
    I’m not sure this is correct. If the central clock receives red-shifted signals and the rotating clock receives blue-shifted signals, then shouldn’t the rotating clock judge the central clock to fun faster than itself, not slower?
     
  17. Jul 21, 2014 #16

    WannabeNewton

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    TumblingDice's statement is incorrect as applied to non-inertial frames. There is no such symmetry in the case of non-inertial frames, it is only existent in general between inertial frames. For starters, the rotating clock circles around the central clock in the latter's rest frame but the central clock is at rest in the rest frame of the former. Secondly, although it is not the case here, naive applications of the time dilation formula to rotating frames can lead to paradoxes. I can describe such a paradox if you wish to see it; it is something I picked up from a famous problem book on relativity.
     
  18. Jul 22, 2014 #17
    Yes, I thought so. By the way, thanks for the earlier reference.

    O.K. We can leave the concept of the energy of light to Albert Einstein and Max Planck.

    I am aware of the contentious nature of the expression “relativistic mass” for bodies which in one definition is given by [itex]M = m/\sqrt{1 - v^2/c^2}[/itex]. The value of this equation is debatable. So again, we shall leave it to the experts.

    I enjoy a good paradox, and a good read...what is the famous problem book?

    I am guessing that since there is no symmetry between an inertial frame and a non-inertial frame, the central clock would see the rotating clock in “slow motion” while the rotating clock would see the central clock in “fast motion”, if it were possible to exchange signals “simultaneously”. Is simultaneous communication possible if the separation between the two bodies is negligible? We know long distance simultaneous communication is lost on earth when allowing for the finite speed of a signal from a satellite, but if the central clock was very close to the rotating clock would this more or less remove the difficulty?
     
    Last edited: Jul 22, 2014
  19. Jul 22, 2014 #18

    WannabeNewton

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    "Problem Book in Relativity and Gravitation"-Lightman et al, problem 1.11. I will paraphrase it below.

    We have two rings that rotate with equal and opposite angular velocity ##\omega## about a common center of a global inertial frame. Adam rides on one ring and Eve on the other and at some initial moment when they pass one another they have their clocks synchronized. For example they could hold in place the arms of their clocks until they pass by one another and at that instant let the clocks start ticking so as to be synchronized at said instant. At this instant, Eve determines Adam's clock to tick slower due to time dilation so when they pass by each other again she will expect Adam's clock to be behind. But the same holds true for Adam as well: he will determine Eve's clock to tick slower at that initial moment so when they meet again he will expect Eve's clock to be behind.

    The questions then are: (1) Is this what really happens? (2) If not, what is the resolution?

    What is "simultaneous communication"?
     
  20. Jul 22, 2014 #19

    Nugatory

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    As long as there is any separation at all between the two bodies, simultaneous communication is impossible: There can be no observer for whom the transmission and the reception of the message happen at the same time; and all observers will agree that the transmission happened before the reception.

    However, if the distance is small enough, we may not be aware of the non-simultaneity. The time it takes for light to cross a modest-sized room is maybe one one-millionth of the smallest time interval that humans can perceive and react to, so the difference between "when signal was sent" and "when signal is received" is so small that we don't notice.
     
  21. Jul 22, 2014 #20
    I agree, Nugatory. Even if we were in the same room having this conversation the communication would be non-simultaneous due to the finite speed of nerve impulses. But it would be “more or less” simultaneous in common parlance. This is what I actually meant. If the separation between clocks is so small as to be neglected, then a hypothetical rotating astronaut could carry on a conversation with a central observer in so called “real time”, if such an imprecise phrase is permitted.

    This is a good one!

    Here is my answer without looking this problem up. Up until now, we had been discussing frequency exchanges between a central (inertial) clock and a rotating (non-inertial) clock to conclude that frequencies were asymmetric. In contrast, two inertial frames consider each other's clock as ticking slower than itself because there is no preferred frame of reference; therefore the frequency exchange is symmetric. (An analogy would be two persons separated by distance judging each other to look small).

    However in the case of Adam and Eve that “instant” resembles the tortoise and the hare paradox. An “instant” is only a mathematical figure of speech which leads to the paradox that the hare can never catch up to the tortoise. So Adam and Eve’s conclusions about each other’s times are wrong. The "instant" cannot be regarded as a snapshot of inertia. Instead, they should assume their respective clocks tick at the same rate i.e. are symmetric.

    P.S. Adam and Eve are constantly moving away and toward each other. Their signals would be Doppler shifted so that communications are always delayed according to the degree of separation. Each other's clock seems to tick slower and slower, or faster and and faster, in alternating sequences. Only when they meet will their clocks appear synchronized for an infitesimal, because as Nugatory pointed out there is no such thing as simultaneity between separated bodies.
     
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